thermodynamics · 2.7 first law of thermodynamics for variable flow process 17 2.8 examples 18 3....

THERMODYNAMICS

For MECHANICAL ENGINEERING

SYLLABUS Thermodynamic systems and processes; properties of pure substances, behaviour of ideal and real gases; zeroth and first laws of thermodynamics, calculation of work and heat in various processes; second law of thermodynamics; thermodynamic property charts and tables, availability and irreversibility; thermodynamic relations. Power Engineering: Air and gas compressors; vapour and gas power cycles, concepts of regeneration and reheat. I.C. Engines: Air-standard Otto, Diesel and dual cycles. Refrigeration and air-conditioning: Vapour and gas refrigeration and heat pump cycles; properties of moist air, psychrometric chart, basic psychrometric processes.

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THERMODYNAMICS

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Topics Page No 1. BASIC CONCEPTS OF THERMODYNAMICS

1.1 Thermodynamic 01 1.2 Microscopic Approach and Macroscopic Approach 01 1.3 System 01 1.4 Property of System 02

1.5 State 03 1.6 Thermodynamic Equilibrium 04

1.7 Zeroth Law of Thermodynamics 04 1.8 Ideal gas Equation and Process 06 1.9 Work Transfer 08 1.10 Heat Transfer 10

1.11 Example 11 2. FIRST LAW OF THERMODYNAMICS

2.1 First Law of Thermodynamics for a Cycle 13 2.2 Application of First Law of Thermodynamics 13 2.3 Meyer’s Formula 14 2.4 Heat Transfer in Different Process 15

2.5 Free Expansion 15 2.6 First Law of Thermodynamics for Flow Process 16

2.7 First Law of Thermodynamics for Variable Flow Process 17 2.8 Examples 18 3. SECOND LAW OF THERMODYNAMICS - ENTROPY

3.1 Thermal Reservoir 23 3.2 Equivalence of Kelvin Plank and Clausius Statement 25 3.3 Carnot Cycle 25 3.4 Refrigeration & Heat Pump working on Reversed Carnot Cycle 26

3.5 Clausius Theorem 26 3.6 Entropy 28

3.7 Combined First Law and Second Law of Thermodynamics 28 3.8 Entropy change for an ideal gas 29 3.9 Entropy change for finite body 29 3.10 Slope of isobaric process & isochoric process on T-S Diagram 29 3.11 Example 29 4. AVAILABLE ENERGY & THERMODYNAMIC RELATIONS

4.1 Available Energy 34

CONTENTS


4.2 Unavailable Energy 35 4.3 Loss in Available Energy 35 4.4 Available Function 35

4.5 Irreversibility 36 4.6 Exact Differential Equations 36

4.7 Maxwell’s Equations 36 4.8 T – DS Equations 36 4.9 Energy Equation 38 4.10 Enthalpy Equation 38 4.11 Difference in Heat Capacities 38 4.12 Joule Kelvin Coefficient 39 4.13 Examples 39 5. PROPERTIES OF PURE SUBSTANCE & GAS MIXTURE

5.1 Pure substance 44 5.2 P – V diagram of Pure substance 44 5.3 Triple Point 45 5.4 Gibbs phase rule 45

5.5 Phase Change of Pure substance 45 5.6 T – S diagram of pure substance 46

5.7 Properties of Pure Substance 47 5.8 Specific volume, Enthalpy and Entropy of different Phases 48 5.9 Equation of State 49 5.10 Properties of Gas mixture 51

5.11 Examples 51

6. REFRIGERATION

6.1 Introduction 56 6.2 Air Standard Refrigeration Cycle 56 6.3 Refrigerator working on reversed brayton Cycle 58 6.4 Vapour Compression Refrigeration Cycle 60

6.5 Effect of Parameters on COP of Vapour Compression Refrigeration cycle 62 6.6 Vapour Absorption Refrigeration Systems 63

6.7 Refrigerant 65 6.8 Designation of refrigerants 66 6.9 Examples 66

7. PSYCHOMETRY

7.1 Properties of Moist Air 72 7.2 Psychometric Chart 73

7.3 Significance, Different Lines on Psychometric Chart 74 7.4 Different Process on Psychometric Chart 78

7.5 Examples 79


8. POWER PLANT ENGINEERING

8.1 Stirling cycle 84 8.2 Ericson cycle 84 8.3 Gas Power Plant 84 8.4 Steam Power Plant 88 8.5 Nozzle and Diffuser 90 8.6 Compressor 92 8.7 Steam turbine 93 8.8 Examples 93

9. INTERNAL COMBUSTION ENGINE

9.1 Heat Engine 101 9.2 Applications of IC Engines 101 9.3 Classifications of IC Engines 101 9.4 Basic components of IC Engine 101 9.5 Terms used in internal combustion engine 102 9.6 Difference between four stroke and two stroke Engine 103 9.7 Performance Parameters 103 9.8 Air standard cycle and efficiency 103 9.9 Comparison of Otto, Diesel, dual Cycle 106 9.10 Examples 107

10. GATE QUESTIONS 112


1.1 THERMODYNAMICS

Thermodynamics is the science of energy transfer and its effect on the physical properties of the substances. Some students have difficulties with thermodynamics because of global nature of its applicability. Most students are used to courses that focus on a few specific topics like statics, dynamics, fluid flows etc. all deals with the limited range of topics. Thermodynamics, on the other hands, deals with many issues that are inherent in every engineering system. A thermodynamic analysis can span from analyzing a complete power plant to analyzing the smallest component in power plant. We begin by introducing some basic thermodynamic terms and definitions. Some of these terms are already in our everyday vocabulary as a result of the broad result of thermodynamics concepts in non engineering concepts (for example cooling process of tea in the open environment is a thermodynamic process).

1.2 MICROSCOPIC APPROACH AND MACROSCOPIC APPROACH

In microscopic approach, a certain quantity of matter is considered with considering the event occurring at molecular level. It is called as statistical Thermodynamics. In Macroscopic approach, a certain quantity of matter is considered without considering the event occurring at molecular level. In this, average behaviour of molecules is considered. It is called as Dynamic Thermodynamic. At the higher altitude or where the density of the system is low, the microscopic approach is used for checking the behaviour of the system.

1.3 SYSTEM

A quantity of matter or region in space upon which attention is concentrated is known as thermodynamic System. The system and the boundary are specified by the analyst, these are not specified in a problem statement.

1.3.1 SURROUNDING

Anything external to the system is called as surrounding or Environment. The combination of the system and surrounding makes universe. It means in the universe if anything is specified by the analyst as a system the things except that system are considered as the surrounding.

Fig.1.1system, surrounding and boundary

1.3.2 BOUNDARY

The separation line which separates the system from surrounding is called as Boundary. Boundary may be fixed or rigid, may be

real or imaginary.

1.3.3 UNIVERSE

The combination of system and surrounding is called as Universe. Universe = system + Surrounding

1.3.4 TYPE OF SYSTEM

There are three types of thermodynamic systems.

1 BASIC CONCEPTS OF THERMODYNAMICS


1) Closed System:

The system in which only energy can transfer without transferring the mass across the system boundary is called as closed system. An example of closed system is mass of gas in the piston cylinder without valve. Let us consider the piston cylinder arrangement as shown in fig 1.2. When heat is supplied to system, gas inside the cylinder expands and thermal energy is converted into mechanical work. But mass of the gas will be the same after the process in cylinder it means there is no mass which crosses the boundary of the system so it is considered as closed system.

fig.:1.2 closed system 2) Open System:

The system in which energy and the mass can transfer across the system boundary is called as open system. Most of engineering devices are open systems. Examples: Turbine, Pump, Boiler, condenser Etc.

Fig. 1.3 open system Let us consider the piston cylinder arrangement with inlet and outlet valve. When mass of a fluid enters in control volume of system, the mass and energy of the system change and then this mass of

fluid id transmitted from the outlet valve (for example the process of compression in compressor),the mass with energy leaves from the system it means in this system mass and energy both crosses the boundary so this is considered as the Open System. 3) Isolated System: The system in which, neither energy nor mass can transfer across the system boundary is called as isolated system. Example: Universe, Insulated thermal flask. Let consider a proper insulated thermal flask and hot water is filled in the thermal flask after passing of the time mass and temperature are same which shows neither energy nor mass crosses the thermal boundary of the flask, so it is considered as an isolated system.

Fig. 1.4 isolated system 1.4 PROPERTY OF SYSTEM A property of the system is a characteristic of system that depends upon the state of the system .as long as state is fixed, property of system is fixed. It does not depend how the state is reached. So properties depend upon end points only. These are point functions and exact differential. Examples: Pressure, temperature, entropy etc. There are two types of properties: (1) Extensive property: These properties depend upon the mass of the system. If the system mass is changed then these types of properties will change.


For example, volume, internal energy, enthalpy, entropy etc.

2) Intensive Property: These properties do not depend upon the mass of the system. For example, specific volume, specific internal energy, specific enthalpy, specific entropy, pressure, Temperature etc. Most of the extensive properties can be converted in to intensive properties by dividing the extensive property by system mass or number of moles in the system. Intensive properties created in this way are called as specific property. It means all the specific properties are intensive property. Let us we an example: we know that volume of the system is considered as the extensive property and its unit is m3.

But Specific volume (v) = V

m . The specific

volume is given by m3/kg. It is given as intensive property. 1.5 STATE It is the condition of a system at the instant of time. Each property has its single value at every state. As long state is fixed the property at that state is fixed. It means that the property does not depend the parameter other than the state or the specific point. Let us consider a state 1 on P- V diagram as shown in fig 1.5. the properties at point 1 are P1 , V1 . The state change from 1 to 2 and the properties at state 2 are P2, V2. then we follow the same points but change the path between the state 1 and state 2 .but the properties remains same. It means properties do not depends upon the path. it only depends upon the point, So result of this discussion is All properties are the point function. Differentiation of all the properties is

exact differential (dT, dP etc.) Properties do not depend upon the past

history. These only depends the present condition of the system

Because in a cycle initial and the final point are same. So for the thermodynamic cycle, a thermodynamic property remains same.

Difference of properties for reversible and irreversible process is same.

1.5.1 PROCESS

A process is said to be done by the system when it change its state. Process may be non flow process in mass does not transfer or it may be flow process in which mass and energy both will transfer. Type of Process: A Process may be reversible process or irreversible process. 1.5.1.1 REVERSIBLE PROCESS

The process is said to be reversible process if when it is reversed in direction it follows the same path as former path without leaving any effect on system and surrounding. Reversible process is most efficient process. 1.5.1.2 IRREVERSIBLE PROCESS

A process which does not follow the condition of the reversible process is called as the irreversible process. The friction is main reason for the process being the irreversible process. 1.5.1.3 QUASISTATIC PROCESS Quasi means ‘almost’ static means ‘slowness’. When process is carried out in very slow manner, it is called as Quasistatic process. Infinite slowness is characteristics of a Quasistatic process. Friction less


quasistatic process is considered as reversible process.

Fig.1.6 quasistatic process Let us consider the expansion of a gas from initial state 1 to final state 2 as shown in fig 1.6. In first case the weight W is release from the system and the gas expands quickly from state 1 to 2. But when weight W is divided into number of parts and then allow to release from the gas in second case. the gas passes through the number of equilibrium state and then reach the final state 2. So second process is very slow process it is known as quasistatic process. The P-V diagram for quasistatic process is shown in fig 1.7.

Fig. 1.7

1.5.2 CYCLE A thermodynamic cycle is defined as a series of state changes such that final state is identical with initial state. for a cycle difference of all the properties is zero.

1.6 THERMODYNAMIC EQUILIBRIUM

A system is said to be in thermodynamic equilibrium when there is no change in the macroscopic property of the system.

A system is said to be in thermodynamic equilibrium when it follows: i) Mechanical Equilibrium ii) Thermal Equilibrium iii)Chemical Equilibrium

i) Mechanical Equilibrium:

If there is no unbalanced force within the system and also between the system and surrounding, the system is said to be in mechanical equilibrium.

ii) Thermal Equilibrium: if temperature of the system is same in all the parts of system, the system is said to be in Thermal equilibrium.

iii) Chemical Equilibrium:

if there is chemical reaction or transfer of matter from one part of system to another, the system is said to be in Chemical equilibrium. 1.7 ZEROTH LAW OF THERMODYNAMICS It states that when body A is in thermal equilibrium with body B is in thermal equilibrium with body C , then body A,B and C will be in thermal equilibrium. It is the basic of temperature measurement. fig. 1.8 Zeroth law of thermodynamics If TA =TB & TB = TC, then TA = TB = TC. 1.7.1 APPLICATION OF ZEROTH LAW The main application of Zeroth law of thermodynamics is to measure the temperature. First thermometric property is measure which change with respect to change of temperature.

TB

TC

TA


Let us consider Mercury in glass thermometer (generally which is used to measure the temperature of human body) as shown in fig 1.9. The length of the Hg change with respected to temperature. So the thermometric property in this type of thermometer is length.

Fig. 1.9 Hg in glass thermometer It is more important to find the change of length so that we can calculate the temperature. So first length of the Hg which is change is found then temperature is calculated. In next topic we explain the process of measurement of the temperature. 1.7.2 TEMPERATURE MEASUREMENT In the thermometer, Zeroth law of thermometer is used to measure the temperature of the body. Thermometric Property: The property which changes with change in temperature is called as thermometric property.

1.7.2.1 THERMOMETRIC PRINCIPLE

The thermometric property which change with the change of temperature is found first and with help of this thermometric property temperature of that specified state is calculated.

1.7.2.2 TYPE OF THERMOMETER

There are five types of thermometers.

a) Resistance thermometer It works on Wheatstone bridge circuit. As temperature hang with respect to change of

temperature resistance change .so thermometric Property in resistance thermometer is resistance. So to find the temperature of any state first resistance of that state is found and then temperature is measured.

Fig. 1.10 Resistance Thermometer (b) Thermocouple This works on See beck Effect. When two dissimilar metal at different temperature is joint with each other, an electro magnetive force is generated the induced e.m.f. depends upon the temperature difference of two ends of the dissimilar metal. So thermometric property in this thermocouple is induced e.m.f.

Fig. 1.11 thermocouple

c) Constant Volume gas thermometer

When the volume in constant volume gas thermometer is constant, the pressure changes with respect to temperature. In this type of thermometers, pressure is considered as the thermometric property.

d) Constant pressure gas thermometer

When the pressure in constant Pressure gas thermometer is constant, the volume


changes with respect to temperature. in this type of thermometers, volume is considered as the thermometric property. e) Mercury –in- glass thermometer

In this type of thermometers, length change with change of the temperature so in mercury in glass thermometer, length is considered as the thermometric property.

1.7.3 METHOD FOR TEMPERATURE MEASUREMENT

1.7.3.1 METHOD FOR TEMPERATURE MEASUREMENT: BEFORE 1954

Before 1954 temperature measurement method, the measurement of temperature is based on two reference point freezing point and boiling point. Let temperature at freezing point is T1 and thermometric property at that temperature is X1 and Temperature at boiling point is T2 and thermometric property at that temperature is X2. Let thermometric Property at temperature T is X .then,

1 1T X

T X ------------- (1)

and Temperature at boiling point is T2 and thermometric property at that temperature is X2. Let thermometric Property at temperature T is X.then,

2 2T X

T X ------------- (2)

From equation (1) and (2)

1 2 1 2T T X X

T X

1 2

1 2

T TT .X

X X

--------- (3)

1.7.3.2 METHOD FOR TEMPERATURE MEASUREMENT: AFTER 1954

The measurement of temperature is based on single reference point triple point of water. Let temperature at triple point is Ti

and thermometric property at that temperature is Xi. Let thermometric Property at temperature T is X. then

Ti = a.Xi --------- (1) and T = a.X ---------(2) Then,

i

XT 372.15

X ----------- (3)

1.7.3.3 TEMPERATURE SCALES There are following temperature scales such as Rankin, Celsius, Kelvin and Fahrenheit. There are two common absolute temperature scales Rankin (R), and Kelvin scale (K). They are related as follows:

9T(R) T(K)

5

Fig. 1.12 different temperature scale The relationship between the Celsius scale and Fahrenheit scale can be given as

0 09T( F) T( C) 32

5 --------- (2)

The relationship between the Celsius scale and Kelvin scale can be given as

0T(K) T( C) 273.15 ------------- (3)

1.8 IDEAL GAS EQUATION AND PROCESS The ideal gas law is the equation of state of a hypothetical ideal gas. It is a good


http://en.wikipedia.org/wiki/Equation_of_state



http://en.wikipedia.org/wiki/Ideal_gas

approximation to the behaviour of many gases under many conditions, although it has several limitations. The ideal gas law is often introduced in its common form:

PV nRT Where P is the absolute pressure of the gas, V is the volume of the gas, n is number of moles of gas, R̅ is universal gas constant, & T is the absolute temperature of the gas. For air R̅ = 8.314 KJ/mol.K n (in moles) is equal to the mass m (in gm) divided by the molar mass M ( in gm/ mole)

mn

M

By replacing n with m / M, we get: m

PV RTM

Defining the specific gas constant R as the ratio R̅/M PV mRT Where R is specific Gas Constant. For air R =0.287 KJ/Kg.K

Different process:

1) Constant volume process An isochoric process, also called a constant-volume process, or an isometric process, is a thermodynamic process during which the volume of the system undergoing such a process remains constant. For isocoric process, V = Constant

1 2

1 2

P P

T T

2) Constant pressure process An isobaric process is a Thermodynamic process in which the pressure stays constant. For isobaric process, P = Constant.

1 2

1 2

V V

T T

3) Isothermal Process: An is other mal process is a change of a system, in which the temperature remains constant. It is also called as hyperbolic process. For Isothermal process, T = Constant or PV = Constant. For the isothermal process,

1 1 2 2P V P V

(4) Adiabatic Process An adiabatic process is a process that occurs without the transfer of heat or matter between a system and its surroundings. Such processes are usually followed or preceded by events that do involve heat transfer. For adiabatic process, PVγ = Constant. P1V1γ = P2V2γ

OR

γ 1

γ2 2

1 1

T P

T P

Adiabatic Process

OR,

γ 1

2 2

1 1

T V

T V


http://en.wikipedia.org/wiki/Gas

http://en.wikipedia.org/wiki/Pressure

http://en.wikipedia.org/wiki/Volume

http://en.wikipedia.org/wiki/Ideal_gas_constant

http://en.wikipedia.org/wiki/Temperature

http://en.wikipedia.org/wiki/Molar_mass

http://en.wikipedia.org/wiki/Gas_constant#Specific_gas_constant

http://en.wikipedia.org/wiki/Thermodynamic_process

http://en.wikipedia.org/wiki/Volume_(thermodynamics)

http://en.wikipedia.org/wiki/Closed_system





http://en.wikipedia.org/wiki/Temperature

http://en.wikipedia.org/wiki/Heat

5) Polytropic Process A polytropic process is a thermodynamic process that obeys the relation:

np C

Where n is polytropic index. P1V1n = P2V2n

OR,

n 1

n2 2

1 1

T P

T P

Polytropic Process

OR ,

n 1

2 2

1 1

T V

T V

Representation of different process on P-V diagram:

kPV C

If k=0, P= C, process is isobaric process. If k = ∞, V =C, process is isocoric process. If k=1 PV=C, process is isothermal process. If k=n PVn =C, process is polytropic process. If k=γ PVγ =C, process is adiabatic process. Different processes on P-V diagram are shown in fig.

Slop of isothermal and adiabatic process on P-V diagram For isothermal process

PV = C Taking log in both the sides logP + logV = log C Differentiating the equation dP dV

0P V

dP dV

P V

dP

dV

P

V

For adiabatic process

PVγ = C Taking log in both the sides log P + γ logV = log C Differentiating the equation dP dV

γ 0P V

dP dVγ

P V

Slopof adiabaticprocessγ

Slopof isothermalprocess

The slop of adiabatic process is higher than the slop of isothermal process.

WORK AND HEAT TRANSFER

1.9 WORK TRANSFER

Work is said to be done by a system if the Sole effect on things external to system can





be reduced to rising of a Weight. The Weight may not be raised, but the net effect external to system would be the raising of a Weight. Let consider the Battery and Motor.

The nature of work changes when boundary between system and Surrounding changes. So work done is boundary phenomenon. It is transient form of energy. Wout Surrounding Win → When Work is done on the system it is

considered as negative. → When work is done by the system, it is

considered as positive. Let Consider a piston cylinder arrangement (i.e. closed).

Work done by the system dW = F. dl = P.A.ds = P.dV

2

1 2

1

W P.dV -------- (1)

It is valid for closed system undergoing a reversible process. So work done by closed system in reversible process is area under p v

diagram about V axis.

If Path A and Path B are two paths but initial and final point is same. Area is

different so work done for different path is different. So work transfer is path function. It is inexact differential. 1.9.1 WORK TRANSFER FOR CLOSED SYSTEM IN VARIOUS REVERSIBLE PROCESSES A) Constant Pressure Process

In the isobaric process pressure remains constant. So work transfer

2

1 2 2 1

1

W p.dv P V V

B) Constant Volume Process

In the isochoric process volume remains constant. So work transfer in this process is zero. V= constant

Work transfer 2

1 2

1

W p.dv 0

C) Isothermal Process

In isothermal process, T= C or

1 1 2 2P.V PV P V C

1 1 2 2P V P VP

V V ---- (1)

2 2

1 2 1 1

1 1

dvW p.dv P V

V

21 1

1

VP V In

V

2 21 2 1 1

1 1

V VW P V In m.RT In

V V

System


D) Adiabatic Process:

In adiabatic Process, γ γ γ

1 1 2 2PV PV P V C

γ γ1 1 2 2

γ γ

P V P VP

V V

2 2 γ

1 21 2 γ

1 1

P VW pdv

V

1- 1-γγ 2 1

1 2 1 1

V VW P V

1

γ 1- γ 1-

2 2 2 1 1 1P V .V P V .V

1 γ

1 1 2 2 1 21 2

P V P V MR(T T )W

1 1

Adiabatic Process

(E) Polytrophic Process

1 1 2 2 1 21 2

P V P V MR(T T )W

n 1 n 1

Where n = Polytrophic index

Polytropic Process

1.10 HEAT TRANSFER Heat is defined as a form of energy that is transferred across the boundary by virtue of a temperature difference. The temperature difference is the potential and heat transfer is the flux. Heat flow into the system is taken as positive and heat flow out of system is taken as negative. The process in which that does not cross the boundary is adiabatic process. Heat transfer is a boundary phenomenon. Heat transfer is transient form of energy. Heat transfer is Path function. For the m mass of substance and ∆T temperature difference, Heat transfer Q α m.∆T Q m.C. T

Where m= mass of substance, ΔT = temperature difference C = Specific heat and its unit is KJ/Kg.K. If m = 1Kg, ΔT = 1 0 C, then C = Q 1.10.1 SPECIFIC HEAT The specific heat of a substance is defined as the amount of heat required to raise the unit mass of substance through a unit rise of temperature. For the gases, in constant volume process it is taken as CV and in constant pressure Process it is taken as Cp. Examples

Q.1 If a gas of Volume 6000 cm3 and at a

pressure of 100 KPa is compressed in reversible Process according to PV2 = C until volume becomes 2000 cm3, determine the final pressure and work done ?

Solution: Given V1 = 6000 cm3 P1 = 100 KPa, V2 = 2000 cm3 P2 =? , 1 2W ?

2 2

1 1 2 2P V P V

100 ×(6000)2 = P2 ×(2000)2


2

2 2

100 6000P

2000

2P 900KPa

Work Transfer

2

1 2

1

W p.dv ----- (i)

In the process 2 2

1 1

2

2 2PV PV P V C

2

CP

V ----- (ii)

Substituting the Value a/p in equation (1)

2

1 2 2

1

CW dv

V

2 1

1 1C

V V

2 2

2 2 1 1

2 1

P V P V

V V

11 22 1 2 PVW P V

=100×(6000)×10-6-900×(2000)×10-6

1 2 1. J W 2K

It means 1.2 KJ work is done on the system. Q.2 Determine the total work done by a

gas system, which follows the expansion Process as shown in Figure.

Solution: Given PA = PB = 50 bar =50 × 105 Pa, VA =0.2 m3, VB = 0.4 m3, VC =0.8 m3,

WA-C = ?

Work done in A-B process = Area under AB

WA-B = 50 × 105 × (0.4 – 0.2) J = 106 J = +1 MJ For the polytrophic process,

n n

B B B CP V P V

50×0.41.3=PB0.81.3 PB = 8.25 bar = 20.3 × 105 Pa

Work done in B-C process = Area under BC on P-V diagram

B B BC

CB

P V

nW

P V

1

5 550 10 0.4 20.3 10 0.8

n 1

B C

61.25 10 JW 1.25MJ

So the total work done B CA BW W 1 1.25 MJW

= 2.25 MJ Q.3 The limiting value of the ratio of the

pressure of gas at the steam point and at the triple point of water when the gas is kept at constant volume is found to be 1.36605. What is the ideal gas temperature of the steam point?

Solution:

t

PGiven : 1.36605

P

T = 273.16 t

P

P

=273.16 × 1.36605

= 373.15 K T = 1000C Q.4 A platinum wire is used as a

resistance thermometer. The wire resistance was found to be 10 ohm and 16 ohm at ice point and steam point respectively, and 30 ohm at sulphur boiling point of 444.6°C. Find the resistance of the wire at 500°C, if the resistance varies with temperature by the relation.

R = R0 (1 +α t + βt2) Solution:


Given: ti=0°C, Ri = 10 ohm, ts =100°C, Rs = 16 ohm For sulphur, tb=444.6°C, Rb=30 ohm, t = 500°C, R=?

10 = R0 (1+0×α+β×02) 16 = R0 (1+100×α + β×1002) 30 = R0 (1 +444.6 ×α + β ×444.62) Solving the equation, we get R= 10 (1 +6.45 ×10-3t+ 4.48×10-6t2) Substituting the value of t = 5000C R= 10 (1 +6.45 ×10-3×500-4.48× 10-6×5002) R = 31.05 ohm. Q.5 The piston of an oil engine, of area

0.0045 m2, moves downwards 75 mm, drawing. In 0.00028 m3 of fresh air from the atmosphere. The pressure in the cylinder is uniform during the process at 80 KPa, while the atmospheric pressure is 101.325 KPa, the difference being due to the flow resistance in the induction pipe and the inlet valve. Estimate the displacement work done by the air finally in the cylinder.

Solution: Given: A=0.0045 m2, length of stroke = 75 mm = 0.075 m

Volume of piston stroke = 0.0045 × 0.075 m3 = 0.0003375 m3

∴ V2-V1 = 0.0003375 m3 As pressure is constant P = 80 KPa So work done = P. (V2-V1)

= 80 ×0.0003375 KJ = 0.027 KJ = 27 J Q.6 A mass of 1.5 kg of air is compressed

in a quasi-static process from 0.1 MPa to 0.7 MPa for which PV = constant. The initial density of air is 1.16 kg/m3. Find the work done by the piston to compress the air.

Solution: Given: m=1.5 kg, P1=0.1 MPa, P2 =0.7 MPa, = 1.16 kg/m3, W=?

For quasi-static process, PV = C

∴ P1V1 = P2V2 = C

311

1

m 1.5V 1.29 m

1.16

Work done

2 11 2 1 1 1 1

1 2

V PW P V In P V In

V P

0.1

0.1 1.29In0.

2 17

.5 MJ

Q.7 680 kg of fish at 5°C are to be frozen

and stored at – 12°C. The specific heat of fish Above freezing point is 3.182 KJ/Kg.K, and below freezing point is 1.717 KJ/Kg K. The freezing point is – 2°C, and the latent heat of fusion is 234.5 KJ/Kg. How much heat must be removed to cool the fish, and what percent of this is latent heat?

Solution: Given: m=680 kg, T1=50C, T2 = -120C, Cp1 = 3.182 KJ/Kg.K, Cp2 = 1.717 KJ /Kg.K, Tf = -20C LH = 234.5 KJ/Kg, Heat to be removed above freezing point

= 680 × 3.182 × {5 – (-2)} kJ = 15.146 MJ

Heat to be removed latent heat = 680 × 234.5 kJ = 159.460 MJ

Heat to be removed below freezing point

= 680 × 1.717 × {– 2 – (– 12)} kJ = 11.676 MJ

∴ Total Heat = 186.2816 MJ Percentage of Latent heat = (Latent heat/total heat) × 100

= (159.460/186.2816) × 100 = 85.6 %


2.1 FIRST LAW OF THERMODYNAMIC FOR A CYCLE

Heat and work are different forms of energy. Both form of energy can be Conserved. As the result of Joule’s experiment, for a closed system undergoing a Cycle (ΣQ) cycle =J (ΣW) cycle

J is Joule’s equivalent. This is first law of thermodynamics for a closed system undergoing a cycle. Both Heat and work are measured in the desired unit of energy Joule (J). So constant of proportionality ‘J’ is unity.

2.2 APPLICATION OF FIRST LAW OF THERMODYNAMICS

2.2.1 HEAT TRANSFER IS A PATH FUNCTION

Let consider a cycle 1-A-2-B-1 and apply first law of thermodynamics for a cycle (ΣQ) cycle =J (ΣW) cycle

A B A B(dQ) (dQ) (δw) (δw) ---- (I)

Let consider another cycle 1-A-2-B-1 and apply the first law of Thermodynamics again

A C A C(dQ) (dQ) (δw) (δw) ------ (II)

Solving the equation (I) or equation (II)

B C B C(dQ) (dQ) (δw) (δw) ------ (III)

Difference B C(δw) (δw) is not Zero

because work transfer is a Path function as it was proved with help of chapter 1, so

B C(dQ) (dQ) is not equal to Zero. So heat

transfer for the different path has the different value. It depends upon the path. So results of this discussion are: Heat is a path function. The differentiation of the heat transfer

is inexact differential (δQ). It is boundary phenomenon. it is transient form of energy.

2.2.2 ENERGY IS A PROPERTY OF SYSTEM

From the equation (III), We get

B C B C(dQ) (dQ) (δw) (δw) ------ (III)

Rearranging the equation (III)

BB CC(δQ) (δW) (δQ) (δw)

B C

(δQ) (δW) (δQ) (δW)

For path B and C, the difference of heat transfer and work transfer is constant. So this different must be a point function and property of the system. This property is known as energy.

Q W dE --------- (IV)

This is known as first law of thermodynamic for closed system undergoing a process. It is valid for the reversible and irreversible process. Total Energy E = Kinetic Energy + Potential Energy + Internal Energy. In absence of motion and gravity, kinetic & potential energy is Zero. In that case E = U δQ ̶ δW = dU ----- (V)

Where, U = Internal energy Unit of internal energy = KJ Internal energy is an extensive

property. This is first Law of thermodynamics for

reversible and irreversible process.But we know that δW=P.dV for areversible process, so

Q P.dV dU ---- (VI)

This is first Law of thermodynamics fora closed System undergoing a reversibleProcess.

2 FIRST LAW OF THERMODYNAMICS


2.2.3 ENERGY OF AN ISOLATED SYSTEM IS CONSTANT According to I Law of thermodynamics δQ = δw + dE For isolated System δQ = 0 δw = 0 dE = 0

1 2E E

Energy of an isolated System is always constant.

Universe is an isolated system so energy of the Universe remains Constant.

It is known as law of conservation of energy.

2.2.4 PERPETUAL MOTION MACHINE (PMM-I) IS NOT POSSIBLE Perpetual machine is the machine which delivers the work without taking any input. There is no such machine which would continuously supply mechanical work without some other form of energy dissipating. Such a machine is called as PMM-I. The efficiency of PMM-1 is given by

PMM I

W Wn

Q O

PMM In

It means efficiency of PMM-1 is Infinite. But according to first law of thermodynamics (ΣQ) cycle = (ΣW) cycle

If it is possible it will violate of first law of thermodynamics but it is law of nature. So PMM-1 is not possible. The machine having the infinite

efficiency is not possible. 2.2.5 ENTHALPY Enthalpy of a Substance is defined as

H = P.V+ U

Enthalpy is an extensive property and its unit is KJ. Enthalpy of an ideal gas is given by H = m.Cp.T

2.2.5.1 SPECIFIC ENTHALPY It is amount of enthalpy per unit mass. Specific enthalpy is an intensive property. Its unit is KJ/Kg. 2.2.6 INTERNAL ENERGY All fluids store energy. The store of energy within any fluid can be increased or decreased as the result of various processes carried out on or by the fluid. The energy stored within the fluid which results from the internal motion of its atom and molecules is called as Internal Energy. The Internal energy of any ideal gas is given:

U = m.Cv.T Internal Energy is an extensive property. It is a point function. Unit of internal energy is KJ. 2.2.6.1 SPECIFIC INTERNAL ENERGY

It is amount of internal energy per unit mass. Specific internal energy is an intensive property. It is denoted by u. Its unit is KJ/Kg. 2.3 MEYER’S FORMULA

We know that Enthalpy can be given as H = U + PV For an ideal gas, H = m.Cp.T, U = m.Cv.T and PV = mRT Substituting the values in equation H = m.Cv.T + mRT m.Cp.T = mT (CV+R)

CP Cp

CP R,C

CV 1

V = R

This formula is known as Meyer’s formula.

We know that p

CP R,C

CV 1


And V

RC

1

2.4 HEAT TRANSFER IN DIFFERENT PROCESS 1) Constant Volume Process: According to first law of thermodynamics δQ = P.dV + dU δQ = O + dU δQ = dU = mCVdT

2) Constant Pressure Process: According to first law of thermodynamics δQ = P.dV + dU We know that for isobaric Process dH = P.dV + dU δQ = dH = m.Cp.dT

3) Isothermal Process: According to first law of thermodynamics δQ = P.dV + dU For isothermal Process dU = 0 δQ = P.dV = δW

δQ = δW = 21 1 n

1

VP V l

V

4) Adiabatic Process:

There is no heat transfer in adiabatic Process. 5) Polytrophic Process: According to first law of thermodynamics δQ= δW + dU

1 2V

mr(T T )mC dT

n 1

1 2 1 2mR(T T ) mR(T T )

1 n 1

1 2

1 1mR(T T )

n 1 1

1 2mR(T T ) n

n 1 1

nQ work done

1

2.5 FREE EXPANSION

The expansion of gas against vacuum is known as free Expansion. There is no work done in free Expansion.

2

1

W 0 although 2

1

P.dv 0


Since work transfer =2

1

P.dv is considered

for reversible Process but free expansion is highly irreversible Process. Vacuum does not offer any resistance, so there is no work transfer involved in free expansion. 2.6 FIRST LAW OF THERMODYNAMICS FOR FLOW PROCESS Steady Flow Process Steady flow means the rate of flow of mass and energy across the control volume are constant. At steady state of a System, any thermodynamic property will have a fixed value at the same state and does not change with respect to time. Flow Work The Flow work is the work done by Fluid of Mass m either at inlet or at the outlet of control Volume. Flow work at inlet is taken as (-P1V1) and at outlet of control volume is (+P1V1).where P1, V1 is pressure and volume at the inlet of control volume and P2,V2 is the pressure and volume at the outlet of control volume. so total work total CV 2 2 1 1W W P V PV

2.6.1 STEADY FLOW ENERGY EQUATION By Appling the energy balance equation at section 1-1 and section 2-2 (Energy at Section 1-1)= (Energy at Section 2-2)

Steady state energy flow [Internal Energy + Kinetic energy + Potential energy + Heat] 1-1

= [Internal Energy + Kinetic Energy + Potential Energy + Work] 2-2

2 2

1 1 1 2 2 2 Total

1 1U mC mgZ Q U mC mgZ w

2 2

2

1 1 1 2

1U mC mgZ Q U

2

2

2 2 cv 1 1 2 2

1mC mgZ w P V P V

2

2

1 1 1 1 1

1U P V mC mgZ Q

2

2

2 2 2 2 2 cv

1U P V mC mgZ w

2

2

1 1 1

1H mC mgZ Q

2

2

2 2 2 cv

1H mC mgZ w

2 Or

2

11 1

2 cv2 2 2

C QH gZ

2 m

w1H C gZ

2 m

It is known as steady Flow energy Equation (SFEE) Where h1,h2 = Specific enthalpy at inlet and outlet KJ/Kg C1,C2 = Velocity at inlet and outlet m/Sec.

cvw = Work done by System KJ

δQ = Heat transfer KJ Z1, Z2 = Elevation above on arbitrary datum , m 2.6.2 APPLICATION OF S.F.E.E. 1) Nozzle For a turbine which is well insulated, and

1 2Z Z

SFEE equation: 2 2

1 21 2 2o o

C CQ Wh1 gZ h gZ

2 m 2 m

Here o o

Q W0

m m


1 2Z Z

2 2

1 21 2

C Ch h

2 2

2 2

2 1 1 2C C 2(h h )

If C1 is very less as Compared to C2, then

2 1 2C 2(h h ) m/s.

2) Turbine For a turbine which is well insulated, the flow Velocities are often small, so the K.E can be neglected and 1 2Z Z

SFEE equation: 2 2

1 21 1 2 2o o

C CQ Wh gZ h gZ

2 m 2 m

cv1 2 o

Wh h

m

0

1 2W m (h h )KJ

3) Compressor

For a turbine which is well insulated, the flow Velocities are often small, so the K.E can be neglected and = SFEE equation:

2 2

1 21 1 2 2o o

C CQ Wh gZ h gZ

2 m 2 m

cv1 2 o

Wh h

m

0

cv 1 2W m (h h )KJ

But in Case of turbine h1 is greater than h2 and in Case of Compressor h2 is greater than h1. 4) Throttling Process When the fluid flows through a constricted passage, like an opened valve or an orifice it is called as throttling Process. Throttling process is irreversible process. In throttling Process,

Q W0

m m

And change of P.E. P.K.E. is very small and ignored, then

2 2

1 21 1 2 2

C CQ Wh gZ h gZ

2 m 2 m

1 2h h

So Enthalpy of fluid before throttling and after throttling remains Constant.

2.7 FIRST LAW OF THERMODYNAMICS FOR VARIABLE FLOW PROCESS Variable Flow If there is a change of property at the same state, Process is called as Variable flow or unsteady Flow Process. In the unsteady flow then there is an accumulation of mass and energy.


The mass Accumulated in control volume is given by

1 2

cv

dm dmdm

dt dt dt

Applying the law of conservation of energy Rate of energy accumulation = Rate of energy inlet Rate of energy outlet

0 2

1 1 1 1 1 1

2

2 2 2 2 2 2

dE d 1cv m h m c m gZ Q

dt dt 2

d 1m h m c m Z W

dt 2

Unsteady flow

0 0 2 0

1 1 1 1 1 1

dE 1 δQcv= m h + m c +m g2 + -

dt 2 δt

0 0 2 0

e 2 e 2 e 2

1 δwm h + m c +m g2 +

2 δt

Neglecting the Kinetic energy and potential energy

0 0 0 0

1 1 e 2

dUcv m h m h Q W

dt

Where 0

lm Mass Flow rate at inlet Kg/Sec 0

em Mass Flow tare at outlet Kg/Sec 0Q Heat transfer Rate KJ/Sec 0W Work done KJ/Sec

1 2h ,h Specific Enthalpy at Inlet or Outlet.

Examples Q.1 A slow chemical reaction takes place

in a fluid at the constant pressure of 0.1 MPa. The fluid is surrounded by a perfect heat insulator during the reaction which begins at state 1 and ends at state 2. The insulation is

then removed and 105 kJ of heat flow to the surroundings as the fluid goes to state 3. The following data are observed for the fluid at states 1, 2 and 3.

State v (m3) t (°C) 1 0.003 20 2 0.3 370 3 0.06 20

For the fluid system, calculate E2 and E3, if E1 = 0

Solution: From first law of thermodynamics δQ = dE +PdV

Q1-2 = (E2 – E1) +

= (E2 – E1) + 0.1× 103 (0.3- 0.003) Because Process 1-2 is insulated so Q1-2 = 0 and E1 = 0

So E2 = - 29.7 KJ

Q2-3 = (E3 – E2) +

-105=(E3+29.7)+0.1× 103 (0.06- 0.3) E3 = -110.7 KJ Q.2 A domestic refrigerator is loaded

with food and the door closed. During a certain period the machine consumes 1 kWh of energy and the internal energy of the system drops by 5000 kJ. Find the net heat transfer for the system. Solution: According to first law of thermodynamics

δQ ̶ δW = dE δQ ̶ δW = dE δW = -1× 3600 = - 3600 KJ. dE = -5000 KJ δQ + 3600 = -5000 δQ = - 8600 KJ = -8.6 MJ Q.3 The properties of a certain fluid are related as follows: u = 196 + 0.718t pv = 0.287 (t + 273)

Where u is the specific internal energy (kJ/kg), t is in °C, p is pressure (KN/m2), and v is specific volume (m3/kg). For this fluid, find Cv and Cp.


Solution:

we know that Cp =P

dh

dT

P

d u pu

dT

P

d 196 0.718t 0.287 t 273

dT

P

0 0.718dt 0.287(dt 0)

dT

P

0.718dt 0.287dt

dT

P

1.005dt

dT

but T = t + 273 so dT = dt Cp =1.005KJ/Kg.K

again V

duCv

dT

V

d(196 0.718t)

dT

V

0.718dt

dT

but T = t + 273 so dT = dt CV =0.718KJ/Kg.K Q.4 A mass of 8 kg gas expands within a

flexible container so that the P-V relationship is PV1.2 = constant. The initial pressure is 1000 KPa and the initial volume is 1 m3. The final pressure is 5 KPa. If specific internal energy of the gas decreases by 40 KJ/kg, find the heat transfer in magnitude and direction.

Solution: For the polytrophic process P1V1n = P2V2n

1000 ×11.2 = 5 ×V21.2

V2 = 82.7 m3

work done 1 1 2 21 2

P V P VW

n 1

1000 1 5 82.7

1.2 1

1 2W 2932.5KJ

dE = - 8 × 40 = - 320 KJ

1 2Q dE W 320 2932.5

Q = 2612.5KJ 2612.5 KJ heat is supplied to the system.

Q.5 The heat capacity at constant

pressure of a certain system is a function of temperature only and may be expressed as

0

p

41.87C 2.093 J / C

t 100

Where t is the temperature of the system in °C. The system is heated while it is maintained at a pressure of 1 atmosphere until its volume increases from 2000 cm3 to 2400 cm3 and its temperature increases from 0°C to 100°C. a) Find the magnitude of heat

interaction. b) How much does the internal

energy of the system increase? Solution:

If T is temperature in K, then t = T-273 and t + 100 = T – 173

373

1 2

273

41.87Q 2.093 dT

T 173

373

273

418.72.093T ln

T 173

= 2.093(373 - 273) +200

ln100

= 209.3 + 41.87ln2 = 238.32 J

2

1 2 2 1

1

Q E – E PdV

E2 – E1 = Q1-2 –P.(V2-V1) =238.32-101.325(0.0024 0.0020)×1000J

= 197.79 J

Q.6 a gas undergoes a thermodynamic cycle consisting of three processes beginning at an initial state where P1 = 1 bar, V1 = 1.5 m3 and U1 = 512 kJ. The processes are as follows: i) Process 1–2: Compression with

PV = constant to P2 = 2 bar, U2 = 690 kJ


ii) Process 2–3: W23 = 0, Q23 = –150 kJ, and

iii) Process 3–1: W31 = +50 kJ. Neglecting KE and PE changes, determine the heat interactions Q12 and Q31.

Solution: Process (1-2): process is isothermal process. for isothermal process,

Q1-2 = (U2-U1)

= (690 – 512) + 1 105 1.5

= (690 – 512) + 1 102 1.5

= 178 – 103.972 = 74.03 KJ

As W2-3 is ZERO so it is constant volume process. As W31 is positive so expansion is done.

For the process (2-3) Q2-3 = W2-3 + (U3-U2) –150 = 0 + (U3 - 690) U3 = 540 KJ Process (3-1) Q3-1=W3-1+ (U1-U3) = 50+(512-540) = 50 -28 = 22 KJ Q.7 A blower handles 1 kg/s of air at

20°C and consumes a power of 15 kW. The inlet and outlet velocities of air are 100 m/s and 150 m/s respectively. Find the exit air temperature, assuming adiabatic conditions. Take Cp of air is 1.005 KJ/Kg-K.

Solution: From S.F.E.E.

2

11 1

cmh m mgZ Q / dt

2

2

2 2 2

1mh mC mgZ W / dt

2

1×h1 + 1× (100)2/2000 + 1×9.8×Z + 0 = 1×h2 + 1× (200)2/2000 + 1×9.8×Z + 15

h2 – h1 = 8.75 Cp (T2-T1) = 8.75 T2 = 20 + 8.7 = 28.7 0C

Q.8 A turbine operates under steady flow conditions, receiving steam at the following state: Pressure 1.2 MPa, temperature 188°C, enthalpy 2785 kJ/kg, velocity 33.3 m/s and elevation 3 m. The steam leaves the turbine at the following state: Pressure 20 KPa, enthalpy 2512 kJ/kg, velocity 100 m/s, and elevation 0 m. Heat is lost to the surroundings at the rate of 0.29 KJ/s. If the rate of steam flow through the turbine is 0.42 kg/s, what is the power output of the turbine in kW?

Solution: P1 = 1.2 MPa t1 = 188°Ch1 = 2785 kJ/kg C1 = 33.3 m/s Z1= 3 m P2 = 20 kPa h2 = 2512 kJ/kg C2 = 100 m/s

dQ/dt = – 0.29 kJ/s dW/dt =?

By SFEE

2

1 11

C gZ Qm(h )

2000 1000 dt

2

2 22

C gZ Wm(h )

2000 1000 dt

2

2

33.3 9.8 30.42(2785 ) 0.29

2000 1000

100 9.8 0 W0.42(2512 )

2000 1000 dt

W

1169.655 1057.14dt

W

112.515kWdt

Q.9 A nozzle is a device for increasing

the velocity of a steadily flowing stream. At the inlet to a certain nozzle, the enthalpy of the fluid passing is 3000 kJ/kg and the


velocity is 60 m/s. At the discharge end, the enthalpy is 2762 KJ/kg. The nozzle is horizontal and there is negligible heat loss from it. a) Find the velocity at exists from

the nozzle. b) If the inlet area is 0.1 m2 and the

specific volume at inlet is 0.187 m3/Kg, find the mass flow rate.

c) If the specific volume at the nozzle exit is 0.498 m3/Kg, find the exit area of the nozzle.

Solution: Velocity at exit from S.F.E.E.

2

1 11

C gZ Qm(h )

1000 1000 dt

2

2 22

C gZ Wm(h )

2000 1000 dt

h1 = 3000 kJ/kg C1 = 60 m/s h2 = 2762 kJ/kg

1 2

W Q0 Z Z

dt dt

2 2

1 11 2

c ch 0 h 0

1000 1000

22

2c60(3000 ) 0 (2762 ) 0

1000 1000

C22 = 602/ 2000 +3000 - 2762 C2 = 692.532 m/ s

Mass flow rate (m) 1 1

1

A C 0.1 60

v 0.187

= 32.1 Kg/sec

Mass flow rate (m) = 2 2

2

A C

v

2A 692.53232.1

0.498

A2 = 8.02 m2

Q.10 A gas flows steadily through a rotary compressor. The gas enters the

compressor at a temperature of 16°C, a pressure of 100 kPa, and an enthalpy of 391.2kJ/ kg. The gas leaves the compressor at a temperature of 245°C, a pressure of 0.6 MPa, and an enthalpy of 534.5kJ/ kg. There is no heat transfer to or from the gas as it flows through the compressor. a) Evaluate the external work done

per unit mass of gas assuming the gas velocities at entry and exit to be negligible.

b) Evaluate the external work done per unit mass of gas when the gas velocity at entry is 80 m/s and that at exit is 160 m/s.

Solution : t1 = 16°C, P1 = 100 kPa, h1 = 391.2 kJ/kg,t2=245°C P2=0.6 MPa=600 kPa

h2 = 534.5 kJ/kg For V1 and V2 negligible and Z1= Z2 So SFEE 1 2m(h 0) 0 m(h 0) W

1 2W 1 (h h ) (391.2 5345)kJ / kg

= –143.3 kJ/kg C1 = 80 m/s; C2 = 160 m/s Then SFEE

2

11

cm(h 0) 0

1000

2

22

cm(h 0) W

1000

280

1 391.2 01000

2160

1 534.5 0 W1000

δW= (–143.3 – 9.6) kJ/kg = –152.9 kJ/kg

Q.11 Air flows steadily at the rate of 0.4 kg/s through an air compressor, entering at 6 m/s with a pressure of 1 bar and a specific volume of 0.85 m3/kg, and leaving at 4.5 m/s with a pressure of 6.9 bar and a specific volume of 0.16 m3/kg. The internal energy of the air leaving is


88 KJ/kg greater than that of the air entering. Cooling water in a jacket surrounding the cylinder absorbs heat from the air at the rate of 59 W. Calculate the power required to drive the compressor and the inlet and outlet cross-sectional areas.

Solution : m1 = 0.4 kg/s,C1 = 6 m/s P1 = 1 bar = 100 KPa v1 = 0.85 m3 /kg m2 = 0.4 kg/s C2 = 4.5 m/s P2 = 6.9 bar = 690 kPa v2 = 0.16 m3 /kg u2 = u1 + 88 kJ/kg

by SFEE

2

11

2

22

c Qm(h 0)

1000 dt

c Wm(h 0)

2000 dt

2

11 1 1

2

22 2 2

c Qm(u P v )

1000 dt

c Wm(u P v )

2000 dt

W

mdt

2 2

1 21 2 1 1 2 2

c c Qu u P v P v )

2000 dt

0.4( 88 85 110.4 0.0076) 0.059

45.357 0.059 = – 45.416 kJ

Mass flow rate 1 1

1

A Cm

V

211

1

m.v 0.4 0.85A 0.0567m

c 6

Mass flow rate 2 2

2

A Cm

V

222

2

m.V 0.4 0.16A 0.01422 m

V 4.5


The first law of thermodynamics states that certain energy is balanced when system undergoes a thermodynamic process. But it does not give any idea whether a particular process is in feasible condition or not. It does not predict the direction of process. It is second law of thermodynamics which predict whether process is in feasible or not with the concept of entropy.

3.1 THERMAL RESERVOIR

A thermal energy reservoir is defined as a large body of infinite heat capacity which is capable absorbing or rejecting the heat without change of temperature. Source: The thermal energy reservoir from which heat can be absorbed without change of temperature is known as Source. Sink: The thermal energy reservoir in which heat can be rejected without change of temperature is known as Sink. There are two statements of second law of thermodynamics

3.1.1 KELVIN – PLANK STATEMENT OF SECOND LAW

It is impossible for any device that operates on a cycle to receive heat from a single reservoir and produce a net amount of work. In other words, no heat engine can have a thermal efficiency of 100%. The machine having 100% efficiency is known as PMM-II. So PMM-II is not possible.

A heat engine that violates the Kelvin‐Planck statement of the second law cannot be built.

3.1.2 CLAUSIUS STATEMENT OF SECOND LAW

It is impossible to construct a device that operates in a cycle and produces transfer of heat from a lower‐temperature body to higher‐temperature body without any external work. If such a device is possible its C.O.P. will be infinite. So Refrigerator having COP infinite is not possible.

A refrigerator that violates the Clausius statement of the second law cannot be built.

3.1.3 HEAT ENGINE

Heat engines convert heat to work. Heat engine is a thermodynamic system operating in a thermodynamic cycle to which heat is transfer and it is converted in to work.

Fig.: Heat Engine 3.1.4 THERMAL EFFICIENCY

3 SECOND LAW OF THERMODYNAMICS, ENTROPY


Thermal efficiency is the index of performance of a heat engine. It is the fraction of the heat input that is converted to the network. Work output Wout = Q1- Q2

Efficiency

1 2 2

1 1 1

Q Q QW1

Q Q Q

-------(3.1)

3.1.5 REFRIGERATOR AND HEAT PUMP The transfer of heat from a low temperature region to a high‐temperature one requires special devices called refrigerators. Refrigerators are cyclic devices, and the working fluids used in the cycles are called refrigerant.

Fig: Refrigerator 3.1.6 CO-EFFICIENT OF PERFORMANCE The index of performance of a refrigerator is given in the term of co-efficient of performance (COP).It is defined as the ratio of desired effect to given input. For the refrigerator Desired effect = Heat rejected from low temperature body (Q2) Work done on the refrigerator W = Q1-Q2

So

R

2QDesised Effect

work inpuCO

t WP

2R

1 2

Q(COP)

Q Q

3.1.7 HEAT PUMP Heat pump is a thermodynamic system operating in a cycle that removes heat from

low temperature body and delivers heat to high temperature body. To accomplish this energy transfer Heat pumps receive the external energy in the form of work or heat.

Fig: Heat Pump 3.1.8 CO-EFFICIENT OF PERFORMANCE The index of performance of a heat pump is given in the term of co-efficient of performance (COP).It is defined as the ratio of desired effect to given input. For the Heat pump desired effect = Heat supplied to high temperature body (Q1) Work done on the refrigerator W = Q1 - Q2

So

1R

QDesired Effect(COP)

workinput W

1HP

1 2

Q(COP)

Q Q

-------(3.3)

Subtracting the equation (3.2) from equation (3.3)

1 2

1 2HP R

COP COP Q Q

1Q Q

HP R(COP) (COP) 1 ------(3.4)

From the equation (3.1) and equation (3.3)

HP

1(COP)

---------(3.5)

Refrigerators and heat pumps are essentially the same devices; they differ in their Objectives only. Refrigerator is to maintain the refrigerated space at a low temperature. On the other hand, a heat pump absorbs heat from a low‐ temperature source and supplies the heat to a warmer medium.


3.2 EQUIVALENCE OF KELVIN PLANK AND CLAUSIUS STATEMENT The two statements of the second law are equivalent. In other words, any device violates the Kelvin‐Planck statement also violates the Clausius statement and vice versa.

Fig:-The violation of the Kelvin‐Planck statement leads to violation of Clausius. 3.3 CARNOT CYCLE The efficiency of a heat‐engine cycle greatly depends on how the individual processes that make up the cycle are executed. The net work (or efficiency) can be maximized by using reversible processes. The best known reversible cycle is the Carnot cycle. Consider a gas in a cylinder‐piston (closed system). The Carnot cycle has four processes: Process (1‐2) Reversible isothermal expansion: The gas expands slowly, doing work on the surroundings. The heat transfers in the Reversible process from the heat source at TH to the gas. Process (2‐3) Reversible adiabatic expansion: The cylinder‐piston is now insulated (adiabatic) and gas continues to expand reversibly. So, the gas is doing work on the

surrounding and as a result of expansion the gas temperature reduces from TH to TL.

Process (3‐4): Reversible isothermal compression: The gas is allowed to exchange heat with a sink at temperature TL as the gas is being slowly compressed. The heat is transferred from the system to the surroundings such that the gas temperature remains constant at TL.

Process (4‐1): Reversible adiabatic compression: The gas temperature is increasing from TL to TH as a result of reversible adiabatic compression. Carnot cycle is the most efficient cycle operating between two specified temperature limits. The efficiency of all reversible heat engines operating between the two same reservoirs are the same. The thermal efficiency of a heat engine (reversible or irreversible) is:

Lth

H H

QW1

Q Q

For the Carnot cycle, it can be shown:

Lth

H

T1

T -------(3.6)

3.3.1 CARNOT THEOREM It states that all heat engines operating between same temperature limit none has a higher efficiency than a reversible engine. The efficiency of an irreversible (real) cycle is always less than the efficiency of the Carnot cycle operating between the same two reservoirs.


If ηth ≤ ηth Carnot, it may be reversible or irreversible heat engine

If ηth ˃ ηth Carnot, Heat engine is not possible.

3.4 REFRIGERATION AND HEAT PUMP WORKING ON REVERSED CARNOT CYCLE A refrigerator or heat pump that operates on the reverse Carnot cycle is called a Carnot Refrigerator, or a Carnot heat pump. The Coefficient of performance of any refrigerator or heat pump (reversible or irreversible) is given by:

L H

H LR

HHP

L

Q Q,COP COP

Q Q Q Q

COP of all reversible refrigerators or heat pumps can be given by:

L H

R HL

PH L H

T TCOP and COP

T T TT

If (COP)R˂ (COP)R,Rev The refrigerator is irreversible Refrigerator If (COP)R= (COP)R,Rev The refrigerator is reversible Refrigerator If (COP)R˃ (COP)R,Rev Refrigerator is not possible.

3.5 CLAUSIUS THEOREM

it states that for a cycle (reversible or irreversible cycle) the cyclic integral of

δQ

T is equal to zero or less than Zero. It

means

Q0

T

-------(3.8)

δQif

T

∮ ˂ 0, the cycle is irreversible cycle.

δQif 0

T

∮ the cycle is reversible cycle.

δQif 0

T

∮ the cycle is not possible.

3.6 ENTROPY 3.6.1 PHYSICAL SIGNIFICANCE OF ENTROPY

Entropy of a system is defined as the measure of degree of molecular disorder or random existing in the system. If higher is the randomness higher will be entropy. When heat is supplied to the system, randomness of the system increases, so the entropy of the system increases. Reversed effect can be measured when heat is removed from the system. 3.6.2 CLAUSIUS INEQUALITY LAW FOR REVERSIBLE CYCLE Consider the cycle shown below composed of two reversible processes A and B. Apply the Clausius inequality for this cycle.

Apply the Clausius inequality for the cycle made of two internally reversible processes:

net

int rev

Q0

T

2 2

1 1int rev int rev

along path A al

net

ong p

et

ath B

nQ Q

T T

Since the quantity (Qnet/T) rev is independent of the path and must be a property, this property is known as entropy S. The entropy change during a process is related to the heat transfer and the temperature of the system. The entropy is given the symbol S (kJ/K), and the specific entropy is s (kJ/kgK). The entropy change during a reversible process is defined as

Rev

QdS

T

2

2 1

1 Rev

QS S

T

--------(3.9)


3.6.3 CLAUSIUS INEQUALITY LAW FOR IRREVERSIBLE CYCLE Consider the cycle 1-A-2-B-1, shown below, where process A is arbitrary that is, it can be either reversible or irreversible, and process B is internally reversible.

A cycle composed of reversible and irreversible processes.

net

int rev

Q0

T

2 2

1 1int rev int rev

along path A along path B

net netQ Q

T T0

The integral along the reversible path, process B, is the entropy change S1 –S2. Therefore

net

2

1 2

1

S S 0Q

T

2

net2 1

1

QS S

T

In general the entropy change during a process is defined as

netQdS

T

-------(3.10)

= holds for the internally reversible process

>holds for the irreversible process.

3.6.4 ENTROPY CHANGE OF SYSTEM IN REVERSIBLE PROCESS i) Heat is added to the system if heat is added to the system, for a reversible process,

dsdQ

T

the net heat transfer is positive so the entropy change will be positive. It means entropy of the system will increase.

ii) Heat is rejected from the system if heat is rejected from the system, for reversible process

dsdQ

T the net heat transfer is negative so

the entropy change will be negative. It means entropy of the system will decrease. iii) No heat Transfer or adiabatic Process If there is no heat transfer between the system and surrounding the net heat transfer will be zero. It means entropy of the system will be constant.

dsdQ

T

dS = 0 S2 = S1 So reversible adiabatic process is known as isentropic process. 3.6.5 ENTROPY CHANGE FOR IRREVERSIBLE PROCESS

2

netsys 2 1

1

Q kJS S S

T K

Here, the inequality is to remind us that the entropy change of a system during an irreversible process is always greater than

δQ

T, called the entropy transfer. That is,

some entropy is generated or created during an irreversible process, and this generation is due entirely to the presence of irreversibility. The entropy generated during a process is called entropy generation and is denoted as Sgen. We can remove the inequality by noting the following

2

netsys 2 1 gen

1

Q kJS S S S

T K


Sgen is always a positive quantity or zero. Its value depends upon the process and thus it is not a property. Sgen is zero for an internally reversible process. And its value increases with the irreversibility. So entropy generation is a path function. Entropy Change of an Isolated System:- For the isolated system the heat transfer is zero.

So, δQ

dST

for the isolated system, δQ = 0

So dS≥0 The total entropy change of an isolated system during a process always increases or, in the limiting case of a reversible process, remains constant. Universe is an isolated system so entropy of universe during a process always increases or, in the limiting case of a reversible process, remains constant. The increase in entropy principle can be summarized as follows:

gen total

0 Irreversibleprocesses

S S 0 Re versibleprocesses

0 Im possibleprocesses

Heat Transfer as the Area under a T-S Curve: For the reversible process, the equation for dS implies that

netQdS

T

netQ TdS

The heat transfer in a reversible process is the differential area under the process curve plotted on the T-S diagram.

2

net

1

Q TdS (3.11)

This is the second law of thermodynamics for a reversible process.

In the above figure, the heat transfer in an internally reversible process is shown as the area under the process curve plotted on the T-S diagram. 3.7 COMBINED FIRST LAW AND SECOND LAW OF THERMODYNAMICS According to first law of thermodynamics δQ = P.dV + dU -----------------(3.12) for the reversible process According to first law of thermodynamics δQ = T.dS ---------------(3.13) for the reversible process From the equation (3.12) and (3.13) T.dS = P.dV + dU ----------- (3.14) it is valid for reversible as well as irreversible process. Equation (3.14) can be written as T.dS = dH – VdP ------(3.15) 3.8 ENTROPY CHANGE FOR AN IDEAL GAS From the Equation (3.14) entropy change is given:

P.dV dUdS

T T

For an ideal gas,

VdU m.C dT aR

ndP m.

T V

then m.R.dV m.CVdT

dSV T


Integrating the equation from state 1 to state 2 2 2 2

V

1 1 1

m.C dTm.R.dVdS

V T

2 22 1 v

1 1

V TS S m.R In m.C .In (3.16)

V T

From the equation (3.15)

dH VdPdS

T T

For an ideal gas,dH = m.CP dT and V m.R

T P

Then Pm.C dTm.R.dPdS

P T

2 2 2

P

1 1 1

m.C dTm.R.dPdS

P T

2 22 1 p

1 1

P TS S m.R In m.C .In (3.17)

P T

Solving the equation (3.16) and (3.17), We get

2 22 1 p v

1 1

V PS S m.C In m.C .In (3.18)

V P

3.9 ENTROPY CHANGE FOR FINITE BODY For the finite body entropy change may be written as 2 2 2

1 1 1

δQ m.c.dTdS

T T

22 1

1

S S m.C.In (3.19)

3.10 SLOP OF ISOBARIC PROCESS AND ISOCHORIC PROCESS ON T-S DIAGRAM

Isobaric and isocoric process

From the Equation (3.14) T.dS = P.dV + dU For the constant volume process, dV = 0 So T.dS = dU = m.CV dT

v v

dT T(3.20)

dS m.C

From the Equation (3.1) T.dS = dH - V.dP For the constant pressure process, dP = 0 So T.dS = dH = m.CP dT

p p

dT T(3.21)

dS m.C

From the equation 3.20 and 3.21 it is cleared that slop of isobaric process is lesser than slop of isocoric process on T – S diagram. Examples Q.1 An ideal gas cycle is represented by

a rectangle on a P-V diagram. If P1

and P2 are the lower and higher pressures and V1 and V2, the smaller and larger volume, respectively. Then a) Calculate the work done per cycle. b) Indicate which parts of the cycle involve heat flow in the gas c) Show that

2 1

2 1 2 1

1

P V

P P V V

If heat capacities are constant.

Solution:

a) w=area of the cycle

2 1 2 1(P P )(V V )

b) Process a-b and b-c Heat absorbed by 1 Mole of gas in one cycle


ab bc V b a P C bQ Q Q C T T C (T T )

Now, 1a b

2

PT T

P and 2 1 bP V RT

2 2 1c b b

1

V P VT T ;T

V R

1 2V b P P

2 1

P VQ C T 1 C T 1

P V

2 1 2 1 2 1P P

2 1

P V P P V VC C

R P V

Ans….c)

2 1 2 1

2 1 2 1 2 1P V

2 1

P P V VW

Q P V P P V VC C

R P V

2 1 2 1

V 1 2 1 P 2 2 1

R P P V V

C V P P C P V V

P V

1 2V P

2 1 2 1

C C

V PC C

V V P P

1 2

2 1 2 1

γ 1

V γP

V V P P

Q.2 A reversible heat engine operates

between two reservoirs at temperatures of 600℃ and 40℃. The engine drives a reversible refrigerator which operates between reservoirs at temperatures at 40℃ and -20℃. The heat transfer to the heat engine is 2000 KJ and the net work output of the combined engine refrigerator plant is 360 KJ. (a)Evaluate the heat transfer to the refrigerant and the net heat transfer to the reservoir at 400C (b) Reconsider (i) given that the efficiency of the heat engine and the cop of the refrigerator are each 40% of the maximum possible values.

(a) Maximum efficiencies of the heat engine cycle

2max

1

T 313n 1 1 1 0.358 0.642

T 873

= 64.2 %

Again 1

1

W

Q= 0.642

1W 0.642 2000 1284KJ

Maximum cop of the refrigerator cycle

3max

2 3

T(cop)

T T

253

4.22313 253

Also 4

2

Qcop 4.22

W

Since 1 2 w 3w w 60KJ

2 1 1284 36w w w 0 924KJ

Q4 = 4.22 × 924 = 3899 KJ

2 1 1 Q Q W 2000 –1284 716 KJ

Heat rejection to wt 400C reservoir

2 3 716 4823 5539 KJQ Q

b) Efficiency of the actual heat engine Cycle

N = 0.4 ×ηmax = 0.4 × 0.642 1 0.4 0.642 2000 5W 13.6 KJ

2 513.6 –360 153.W 6 KJ

Cop of the actual refrigerator as do. COP = 0.4 × 4.22 = 1.69

Therefore

4 153.6 1.69 259. KJQ 6

3 259.6 153.6 413 2KJQ .


2 1 1 2000 – 513.6 148Q Q 6. KW 4 J

Heat rejected the 400C reservoir

2 3 413.2 1486.4 1=Q 89Q 9.6 KJ

Q.3 A System has a heat capacity at

constant volume CV = AT2 Where A=0.042J/K3. The system is originally at 200 K , and a thermal reservoir at 100 K is available. What is the maximum amount of work that can be recovered as the system is cooled down to the temp of the reservoir?

Solution Heat removed from the system

2 2

1 1

T T 100K

2

V

T T 200K

C dT 0.042 TQ T d

3 (100K)

(200K)0.042[T / 3]

3 3 3 30.042J / K (100 200 )K

3

398 10 J

100K 100K

2

V

200K 200K

dT dTS System C 0.042T

T T

0.042

2

3 2 2 2J / K 100 200 K 630J / K

3

1Q W 98X10 WS

Tresres=

100

S Working flued in HE = 0

S = S system = S ros

398X10 W

60

3010

Since S 0

398 10 W

630 0100

W

980 630 0100

W

350100

W (max) = 35000 J = 35 KJ Q.4 A body of constant heat capacity CP

and initial temperature T1 is placed in contact with a heat reservoir at temperature T2 and comes to thermal equilibrium with it. If

2 1T T , Calculate the entropy

change of the universe and show that it is always positive.

Solution: 2

1

T

2CP CP n

1T

TdTm m l

T TS body

2 1

CP

2

S reservoiT T

mrT

2 1CP

1 2

T Tm In 1

T TS universe=

Putting 1 2

2 1

T T1-

1=n, =

T T 1-n

Since 1

2

T>1

T, n is positive

2 3 4

n

CP

Suniverse n x nln1 l 1 n n

m 2 3 4

(ds) > 0 It means entropy of universe is always positive.

Q.5 A hypothetical device is supplied

with 2 kg/s of air at 4 bar, 300 K. Two separate streams of air leave the device, as shown in figure below. Each stream is at ambient pressures of 1bar, and the mass flow rate is the same for both streams. One of the exits streams is said to be at 330 K while the other is at 270 K. The ambient temperature is 300 K.


Determine whether such a device is possible.

Solution: The entropy generation rate for the control Volume

gen

e e i iS m S mS

2 2 3 3 1 1m S m S m S

2 2 3 3 2 3 1m S m S m m S

2 2 1 3 3 1m S S m S S

Now 2 22 1 P

1 1

T PS S C In R In

T P

330 10.281.005 In 7In 0.494KJ / kg.K

300 4

3 33 1 P

1 1

T PS S C In RIn

T P

270 10.287In

31.005 In

00 4

0.294 KJ / Kg.K

1 0.484 1 0.292 sgen0.786 / KJ K

= 0.786 K Since

Sgen >0, the device is possible.

Such device actually exist and

called Vortex tube.

Q.6 A mass of m kg of fluid at temperate

me T1 is mixed with an equal mass of game fluid at temperature𝑇2 . Prove that the resultant change of entropy of universe is

1 2

1 2

22ME

.

n

T T

lT T

Where C = Specific heat of fluid. Solution: Heat given by part-1 = heat taken by part-2

1 2mc T T mc T T

1 2T TT

2

Change of entropy 1 2ds (dS) (dS)

1 2

T T

T T

dT dTmc

T Tmc

1 2

T Tmcln mcln

T T

22

1 2 1 2

T Tmcln mcln

T .T T .T

But 1 2TT T

2

Substituting the value of T in the Expression

1 2

1 2

T T

2dS 2mc In

T .T

Q.7 A heat pump operates between two

identical bodies. In the beginning , both the bodies are at same temp T1 but operation of heat pump cools down one of the body to temperatureT2. Show that for the operation of heat pump the minimum work input needed by the heat pump for unit mass if given


2

1min 2 1

2

TW T 2T

T

. Where C is

specific heat of bodies. Solution:

Let the final temperature attained by body to which heat is rejected is

fT

W = 𝑄𝐵 − 𝑄𝐴 Heat transfer from body A,

1A 2C(TQ T )

Heat absorbed by body B,

B f 1Q T C T

Work input =

B A f 2 1Q Q C T T 2T

Now entropy change of body A

2

A1

TdS

TCIn

Entropy change of body B

f

B1

TdS

TCIn

For a reversible cycle entropy change of universe = 0.

universe

dS 0

A B

dS dS 0

substituting the values, we get C

2 f

1 1

T Tln ln Cln 0

T T

2 f

2

1

T .TCln 0

T

1 2 fT T .T

2

1f

2

TT

T

Substituting the value of Tf in equation (i)

2

1min 2 1

2

TW C T 2T

T

Q.8 A finite thermal system having heat

capacity C=0.04T2J/K initially at 6oo K. Estimate the maximum work obtained from the thermal system if surrounding temperature is 300 K.

Solution: Q = C dT

300

2

600

Q 0.04T dT

3003

600

0.04T

3

3 30.04300 600

3

= -2520 kJ Now the change of entropy of the body

2 2 2

1 1 1

T

body

T T 2

T T T

ddQ CdT 0.04T d

sT

T T T

2 20.04300 600 5400J / k

2

Entropy change of sink

2

20 0

Q Q WdS

T T

For maximum work considering

dS Universe = 0

1 2

dS dS 0

or 0

Q W5400 0

T

or 2520 W

5400 0300

so W = 900 kJ


The second law of thermodynamics tells us that it is not possible to convert all the heat absorbed by the system in to the work. Only some fraction of heat can be converted on the work. This work will be more valuable as compare to the heat. So Work is high grade energy and heat is the low grade energy.

Example: High grade energy:

1) Mechanical work2) electrical energy3) water power4) wind power5) kinetic energy of a jet6) tidal power.

Example: Low grade energy: 1) Heat or thermal energy2) heat derived from nuclear fission or

fusion.3) Heat derived from combustion of fossil

fuels.4) Solar energy.

4.1 AVAILABLE ENERGY

The maximum amount of work that can be obtained by a system in a cycle is called as Available Energy (AE). Suppose a certain quantity of energy Q1 as heat can be received from a reservoir, at temperature T1.The sink temperature is T2. It rejects the heat to the surrounding at temperature of T2. The maximum possible work done by the system in the cycle is known as Available Energy. The maximum possible work can be achieved when the cycle is reversible cycle.

Fig: heat engine

The maximum work done obtained in the cycle will be the work done in the reversible cycle. So for the reversible cycle,

2th

1 1

TW1

Q T

21 th 1

1

TW Q . Q . 1

T

But the ambient temperature is T0. The sink temperature will be T0.

1 10

0.

1

Q T dT

W Q . 1T

Smax

Where dS is the entropy change of the system. The Carnot cycle and the available energy are shown in figure. The area 1-2-3-4 represents the available energy. Suppose a finite body is used as a source. Let a large number of differential Carnot engines be used with the given body as the source.

fig.: Available and unavailable Energy

Suppose a finite body is used as a source. Let a large number of differential Carnot engines be used with the given body as the source.

0TdW dQ dQ 1

T

If the initial and final temperatures of the source are T1 and T2 respectively, the total

4 AVAILABLE ENERGY & THERMODYNAMIC RELATIONS


work done or the available energy is given by

2 2

1 1

T T

0o

T T

T dQw dQ dQ 1 Q T

T ET

0or w Q T | S|

So

Available Energy 10

1

max

T1AE W Q .

T

= Q1 –T0. (dS) ------(4.1) 4.2 UNAVAILABLE ENERGY The shaded area 4-3-B-A represents the energy, which is discarded to the ambient atmosphere and this quantity of energy, cannot be converted into work and is called Unavailable energy.

o0. 1

1

TUAE T dS Q .

T

----- (4.2)

4.3 LOSS IN AVAILABLE ENERGY Suppose a certain quantity of energy Q is transferred from a body at constant temperature T1 to another body at constant temperature T2 (T2<T1). Initial available energy, with the body at T1, Available Energy

1

11

0max

T1

TAE W Q .

= Q1 –T0. (dS)

Available energy, with the body at T2, Available Energy

!max 1 1 02

0

2

T1

TAE W Q . Q – T . dS

fig.: loss of available energy Decrease in available Energy

0 0

21

1

12

T T

T TAE AE Q .

2

2

11

0

1

T TT

T TQ .

.

= T0. (dS) – T0. (dS!) Decrease in available energy or increase in unavailable energy

1 20

1 2

1

1 0 0=Q . T . dS – T . dST -T

TT .T

----- (4.3)

From equation 3 it clears that when same amount of heat is supplied at the lower temperature its value is low, because it produces the less amount of work as compare to work done produced by the heat when it is supplied by higher temperature. So according to first law of thermodynamics heat at higher and lower temperature has the same significance. But according to second law of thermodynamics heat at the higher temperature is more valuable as compare to heat at lower temperature. So, First law of thermodynamic is called as

quantitative law. Second law of thermodynamic is called

as qualitative law. 4.4 AVAILABILITY FUNCTION The availability of a given system is defined as the maximum useful work that can be obtained in a process in which the system comes to equilibrium with the surroundings or attains the dead state. It is given by maximum work done – atmospheric work. A) AVAILABILITY FUNCTION FOR NON-FLOW PROCESS Let P0 be the ambient pressure, V1 and V0 be the initial and final volumes of the system respectively. If in a process, the system comes into equilibrium with the


surroundings, the work done in pushing back the ambient atmosphere is P0 (V0-V1).

Availability useful max 0 0 1W W –P V V .

Consider a system which interacts with the ambient at T0. Then from the first law of thermodynamics δQ ̶ δW = Du Max work can be obtained when process is reversible process. Let S1 and S2 be the entropies of the systems at the entrance and exit of the device then (dS)system= S2–S1. and (dS) surr = - (Q / To) From the entropy principle

2 1S –S – Q / To 0.

For the reversible process,

2 1S –S – Q / To 0

Therefore 2 1Q To S S

max 1 0 0 1 0W U U T S S

1 0 1 0 0 0U TS – U TS

So Availability = Wuseful = (U1-T0 S1) – (U0-T0 S0)- P0(V0-V1) = (U1+ P0V1-T0 S1) – (U0+P0V0-T0 S0) = ɸ1- ɸ0

Availability Function (=U+P0V-T0S---- (4.4) Where ɸ = U+P0V-T0S is called the availability function for the non flow process. Thus, the availability on any given state: ɸ 1- ɸ 0

If a system undergoes a change of state from the initial state 1 (where the availability is (ɸ 1- ɸ 0) to the final state 2 (where the availability is (ɸ 2- ɸ 0), the change in the availability or the change in maximum useful work associated with the process, is ɸ 1- ɸ2. So the work done by the system when system undergoes from state 1 to state 2 is given by

1 2 1 2 1 0 1 0 1 2 0 2 0 2W f f U P V T S U P V T S

-------- (4.5) B) AVAILABILITY FUNCTION FOR FLOW PROCESS

The maximum power that can be obtained in a steady flow process while the control volume exchanges energy as heat with the ambient at T0: The steady stet steady flow equation is given by

2 2

1 1 1 2 2 2 cv

1 1H mC mgZ Q H mC mgZ w

2 2

The subscript 1 and 2 refer to the entrance and exit respectively. At the exit let the system be in equilibrium with the environment at pressure Po and temperature To. Let symbols without subscripts refer to the entrance condition of the system and changes in KE and PE are negligible. Then useful work is given by W = (H2-H1) + Q The greater the value of Q larger will be the useful work. Thus W will be maximum when Q is a maximum. Let S1 and S2 be the entropies of the systems at the entrance and exit of the device then (dS)system = S2 – S1. and (dS)surr = - (Q / To) From the entropy principle ( S2 –S1 ) – (Q /To) ≥ 0 For the reversible process, ( S2 –S1 ) – (Q /To) = 0 Therefore Q = To (S2- S1). The useful work Wuseful = W = (H2-H1) + To (So-S) = (H1-ToS1) – (H2 – ToS1) So the maximum possible work done Wmax = ψ1 - ψ2 Where ψ is known as the availability function for steady flow. ψ = H - ToS--------- (4.6)

4.5 IRREVERSIBILITY The actual work done by a system is always less than the idealized reversible work, and the difference between the ideal work and actual work is called the irreversibility of the process. So irreversibility can be given as I = Wmax – Wact

This is also some time referred to as degradation or dissipation. For a non flow process between the equilibrium states,


when the system exchanges heat only with the environment

system surr

I U1 U2 –To S1 S2 – U1 U2 Q

= To (S2-S1) – Q = To (dS) system + To (dS)surr = To [(dS)system + (dS)surr ] Hence I> 0. Similarly for the steady flow process I = Wmax – Wact

= To (S2-S1) – Q = To (dS) system + To (dS)surr = To[ (dS)system + (dS)surr ] Thus same expression for irreversibility applies to both flow and non-flow process. The quantity To[ (dS)system + (dS)surr ] represents irreversibility. I = To(dS)universe = To[ (dS)system + (dS)surr ]

-------- (4.7) THERMODYNAMIC RELATIONS 4.6 EXACT DIFFERENTIAL EQUATIONS Theorem No. 1: If a variable Z is a function of the two variable x and y i.e. If Z = f (x , y ) dZ = Mdx + Ndy M and N are the function of x and y. The differential is exact differential if

yx

M N

y x

---------- (4.8)

Theorem No. 2: if a variable f is the function of the x, y, z and a relation exists i.e. If f = f (x, y,z )

f ff

x y z1

y z x

----------- (4.9)

Theorem No. 3: If a variable z is the function of x, y i.e. If x = f (y, z ) ------- (4.10) 4.7 MAXWELL’S EQUATIONS A pure substance in a single phase has only two independent variables. According to combined first Law & Second Law of thermodynamics i) dU = Tds – PdV

ii) dH = Tds + VdP Gibbs function (G) and the Helmholtz function (F) can be given as dG = dH – TdS Gibbs function dF = dU –TdS Helmholtz function Or these functions can be written as iii) dF=-PdV - SdT − Helmholtz function iv) dG = VdP - SdT - Gibbs function U, H, F and G are thermodynamic properties and exact differential ,so Appling the Exact differential rule equation, we get Maxwell equations:

s v

T P

V S

------------ (4.11)

s P

T V

P S

------------ (4.12)

V T

P S

T V

----------- (4.13)

P T

V S

T P

------------ (4.14)

These four equations are known as Maxwell’s equations. 4.8 T – DS EQUATIONS Let Entropy is function of temperature and volume S = f (T , V) Differentiating the equation

V T

S SdS dT dV

T V

multiplying the equation with temperature T

V T

S STds T dT T dV

T V

We know that

V

V

ST C

T

and

T V

S P

V T

substituting the values in TdS equation, we get

v

v

pTds C dT T dv

T

------------(4.15)

This is known as first T.ds Equation.


Let the entropy is the function of temperature T and Pressure P S = f (T , P) Differentiating the equation

P T

S Sds dT dP

T P

multiplying the equation with temperature

P T

S STds T dT T dP

T P

We know that,

P

P T P

S S VT C and

T P T

substituting the values in TdS equation, we get

p

p

vTdS C dT T dp

T

--------(4.16)

This is known as Second T -dS Equation 4.9 ENERGY EQUATION From combined first Law & second Law of thermodynamics dU = TdS – P.Dv Substituting the first TdS equation in the internal energy equation,

V

V

PC dT T dV P.dV

TdU

or

v

v

pdU C dT T p dv (4.17)

T

For an ideal gas PV = mRT

or V

P mRTT P

T V

substituting the value in equation 4.17, we get

VdU C dT P P dV (4.18)

dU =CV.dT

4.10 ENTHALPY EQUATION From combined first and second Law of thermodynamics, dH = TdS + VdP

P

P

VTds C dT T dP

T

P

P

Vd C dT T dP Vdp

T

p

p

vdH C dT V T dP

T

------- (4.19)

For an ideal gas

PV = mRT

P

V mRTT V

T P

Substituting the Value in equation 4.19, we get

PdH C dT V V dP

dH = Cv/dT---- (4.20) 4.11 DIFFERENCE IN HEAT CAPACITIES Equating the first and second T- dS Equation from equation 4.15 and 4.16

P V

P V

V PTdS=C dT T dP C dT T dV

T T

Or P V

V P

P VC C dT T dV T dP

T T

Or

V P

P V P V

P VT dV T dP

T TdT

C C C C

Or

P V

T TdT dV dP

V P

V P

P VP V P V

P VT T

T TT Tand

C C V C C P

------------ (4.21) From the equation 4.21

P V

V P

P VC C T

T T

But V P T

P T V1

T V P

2

P V

TT

V PC C T

T V

------ (4.22)


It indicates the following facts

(a) Since2

V

T

is always positive and

T

P

V

is negative so P VC C is always

positive. (b) at absolute temperature (T=OK);

P VC C

(C ) for an ideal gas

VP mRT

P

V MR V

T P T

2

T

P MRT

V V

2

P V 2 2

V MRTC C T X X

T V

P VC C R ------- (4.23)

4.11.1 VOLUME EXPANSIBILITY β

p

1 V(4.24)

V T

4.11.2 ISOTHERMAL COMPRESSIBILITY

1 UK T (4.25)

V P

substituting the values from equation 4.24 and 4.25 in equation 4.22,

2

P

P V

T

1 VTV

V TC C

1 V

V P

2

P V

TVBC C

K4.26

4.12 JOULE KELVIN COEFFICIENT The slop of an isenthalpic Curve (throttling process) on T-P diagram is known as Joule Kelvin Co-efficient (μ).

h

T

P

------------ (4.27)

A gas is made to undergo continuous throttling process by a valve as shown in fig. Let P1, T1 be the arbitrarily chosen pressure and temperature before throttling and P2, T2 are the pressure and temperature after the throttling. These are plotted on T-P diagram.

Then the initial temperature and pressure of the gas are set to new values, a family of isenthalpic is obtained for the gas as shown in fig. the locus of the points where μ is zero is known as Inversion curve. The region inside the inversion curve when μ is negative is called as heating region. The region inside the inversion curve when μ is positive is called as cooling region.

The enthalpy change of a gas is given by

p

p

VT V dPdh C dT

T

dh = 0

h p

T 1 VT V (4.28)

P Cp T

For an ideal gas,

p

VT V 0

T

So μ=0 So there are some important points:


For an ideal gas Joule Kelvin Co- efficient is Zero.

In the cooling region the value of Joule Kelvin Co efficient is positive.

In the heating region the value of Joule Kelvin Co efficient is negative.

At the inversion curve, value of Joule Kelvin Co efficient is zero.

Examples Q.1 In a steam generator, water is

evaporated at 026.0 c while the

combustion gas P

KJ(C 1.08 .k)

kg is

cooled from 01300 C to 0320 c . The

surrounding are at 030.0 c . Determine the loss of available energy due to the above heat transfer per kg of water evaporated (Latent heat of vaporization at

water at 0260 c 1662.5KJ / kg ).

Solution: Availability decrease of gas

1 1 0 1 2 0 2AE H T S (H T S )

1 2 0 1 2H H T (S S )

2g p 1 2 0 g p

1

Tm c T T T m c ln

T

2g p 1 2 0

1

Tm c T T T ln

T

Given

1 2

0

T 1573k,T 320 273

593kT 30 273 303k

1 g

593AE m X1.08 1573 593 303ln

1573

g739.16m KJ

Availability increase of water

2 1 0AE Q T ds

1 0Tds T ds

1 0T T ds

1 0

1

LHT T m

T

303

1 1662.5 1533

717.4KJ For m mass flow rate of gas Heat given by gas=heat gain by water

g pg 2 1 wm c T T m LH

gm X1.08 1300 320 1X1662.5

gm 1.57kg

1AE 1.57 739.16 1161.1KJ

Loss of Availability = Agas - Aw = (1161.1-717.4) = 443.67 KJ

Q.2 Air expands through a turbine from

500 Kpa, 5200C to 100 KPa , 3000C . During expansion 10 KJ/kg of heat is lost to surrounding which is at 98 KPa , 200C. neglecting K.E. and P.E. changes, determine per kg of air (a) the decrease in availability (b) maximum work (c) the Irreversibility. For air, take

pp 1.005KJ / kg.k,hc c T, where

Cp is constant. Solution: The change of entropy of the air is

2 22 1 p

1 1

T PS S mc ln mR ln

T P

For unit mass of air

2 1 n

513 1S S 1 1.005l ( ) 1 0.287ln

793 5

0.4619 0.3267 0.1352KJ / kg k

Change in availability 1 2 1 0 1 2 0 2 1 2 0 1 2H T S H T S H H T (S S )

P 1 2 0 2 1mC T T T (S S )

520 –300 293 0.1352 261 01 .7 KJ /.0 kg05

Maximum work done

max 1 2w 260.7KJ / kg

But from S.F.E.E

1 2h hQ w

[Neglecting K.E. and P.E.] actual 1 2w Q (h h )

10 1X1.005(520 300)


1 2Q (h h )

211.1 KJ/kg

So, Irreversibility max actualI w w

260.7 211.1 49.6KJ / kg

Q.3 0.2 kg of air at 0300 c is heated reversible at constant pressure to 2066 k. Find the available and unavailable energies of the heat

added. Take 0

OT 30 c and

pc 1.0017 KJ/kg.k.

Solution: Let the 0.2 kg of air is heated at cons leant pressure. So entropy change

2 22 1 P n n

1 1

T PS S mC l mRl

T P

n

20660.2 1.0047l

573

0.2577KJ / K Increase in available energy of air

2 1 0 2 1AE H H T S S

P 2 1mC T T 303 0.2577

0.2X1.0047 2066 573 78.084

1250.14 78.08 1172.2KJ

Heat Input P 2 1mC (T T )

0.2 1.0047(2066 573)

1250.24

Unavailable Energy=Heat Supplied

Available energy 1250.24 1172 78.08KJ Q.4 A pressure vessel has a volume of

31m and contains air at 1.4mPa , 0175 c . The air is coaled to 025 c

by heat transfer to surrounding at 025 c . Calculate the availability in

the initial and final states and the irreversibility of this process. Take

0 aP 100KP .

Solution:

Let mass of air in the vessel is m kg

Mass of air 1 1

1

P V(m)

RT

1400 1

0.287 998

10.89kg

Volume of gas after reaching the dead state

00

0

mRTV

P

10.89 287 298

100

3

0V 9.31m

Availability at initial state 1 1 2AE

1 0 0 1 0 0 1 0U U T S S P (V V )

V 1 2 0

1 1P V 0 1 0

0 0

mC T T T

V PmC ln mC ln P V V

V P

m0.718X 448 298 298

1 14001.005ln 0.718ln

931 100

+ 100(1-9.33)

9.31 150 103.52 831

Q.5 A heat source at 0627 c transfer heat at rate of 3000 KJ/min. To a system

maintained at 0287 c . A heat link is

available at 027 c . Assuming these temperatures to remain constant Find: i) Change in entropy of source ii) Entropy production accompanying

heat transfer iii) The original available energy iv) The Available energy after heat

transfer Solution: Temperature of source

0

1T 627 c 627 273 900K

Temperature of a system 0

2T 287 c 287 273 560K

Sink temperature


0

0T 27 c 27 273 300K

i) Change of entropy of the source

1

1

Q 3000ds 0.056KJ / Ksec

T 900X60

ii) Entropy production accompanying heat transfer heat transfer by

source =Heat absorbed by System

'

1 2Tds T .ds

' 1

2

T .dsds

T

3000 KJ0.089

560X60 Ksec

iii) The original available energy

01

1

T 300Q 1 3000 1

T 900

2000KJ / min iv) Available energy after heat

transfer

01

2

T 300Q 1 3000 1

T 560

1392.8KJ / min Q.6 4 kg of water at 400c mixed with 6

kg of water at 1000c in steady flow process. Calculate: (i) Temperature a resulting mixture (ii) Change in entropy (iii) Unavailable energy with respect

to energy receiving water at 100c .

Solution: (i) The final temperature at the

mixture is Tf , then Heat transfer by 4 kg at water =

Heat transfer by 6 kg at water

1 f 1 2 2 fm c T T m c T T

f f4 4.18 T 40 6 9.18 100 T

f fT 40 1.5 100 T

f fT 1.5T 150 10

0

fT T6 c

(ii) Entropy change of mixture Ds=Entropy change of 4 kg of

water+Entropy change of b kg at water

f f1 p 2 p

1 2

T Tm c ln m c ln

T T

349 3494 4.18ln 6 4.18ln

313 373

1.82 1.66 0.15KJ / K (iii)Unavailable energy with respect

to water at 400c

0UAE T ds system

=29891.82)=542.36KJ

Q.7 Using Maxwell’s relations, show that

for a pure substance

P βTdS C dT TV dV

Where β is co-efficient of thermal expansion, K is co-efficient of compressibility and P VC ,C are

specific heat at constant pressure and constant volume respectively.

Solution: Let entropy of the pure substance S f (P,T)

T P

T STdS T dβ T dT

P T

_ (1)

But

P P

T TSlopof Isobaricprocess

S C

So P

P

ST C

T

And from Maxwell’s equation

T P

S V

P T

Substituting the values in equation(1)

P

P

VTdS C dT T dP

T

Coefficient of thermal expansion

P

1 VB

V T

PTdS=C dT-TVβdP


And entropy of pure substance again S f (T,V)

V T

S SdS dT dV

T V

V T

S STdS T dT dV 2

T V

V V

T TSlopof isocoricprocess

S C

V

V

ST C

T

And from Maxwell’s Relation

V V

S P

T T


V

V

PTds C dT T dV (B)

T

And

T P

1 V 1 VK andβ

V P V T

T P

B P V.

K V T

From cyclic property

T P V

P V T. 1

V T P

T P V

P V P.

V T T

So V

B P

K T


V

BTdS C dT T dV

K

Q.8 Derive the expression for (dh)T for a

substance that obeys the equation of state given by

2

RT qP

V V

Solution: We know that internal energy of pure substance

V

V

PdU C dT T P dV

T

If U f T,V

V T

U UdU dT dV

T V

V

v T v

U U P C , T P

T T T

For isothermal process

T

dU 0

T V

U PT P

V T

But from the relation

2

RT qP

V V

V

P R0

T V

V

P TRT

T V

2

V

P RT qT P P

T V V

So 2

T

U q

V V

2T

qdU dV

V

Integrating the equation

2 2

2

1 1

qdU dV

V

2 1

1 2

1 1U U q

V V

2 1 2 2 1 1 2 1Tdh U U P V PV h h

2 2 1 1

1 2

1 1P V P V q

V V

2 2 1 1Th2 – h1 P V – PV q

1 2

1 1

V V


5.1 PURE SUBSTANCE

A substance that has a fixed chemical composition throughout is called pure substance. Examples-Water, helium carbon dioxide, nitrogen Etc. It does not have to be a single chemical element just as long as it is homogeneous throughout, like air. A mixture of phases of two or more substance is can still a pure substance if it is homogeneous, like ice and water or water and steam.

(Not a pure substance because the composition of liquid air is different)

5.1.1 PHASE OF PURE SUBSTANCES

There are three principle phases – solid, liquid and gas, but a substance can have several other phases within the principle phase. Nevertheless, thermodynamics deals with the primary phases only. In general Solids have strongest molecular bonds.

Their molecules do not move relative toeach other.

Liquid molecular spacing is comparableto solids but their molecules can floatabout in groups.

Gas molecules have weakest molecularbond strength. Molecules in the gasphases are far apart, they have noordered structure. The molecules moverandomly and collide with each other.

5.2 P- V DIAGRAM OF PURE SUBSTANCE

Let the ice be heated slowly so that its temperature is always uniform.fig show the state change of a pure substance (other than water which contract due to conversion from ice to water).line passing through all the saturated solid states is called the saturated solid line. The line passing all the saturated liquid points is called as the saturated liquid line. The line passing all the saturated Vapour points is called as the saturated Vapour line. Where the two lines (saturated liquid line and saturated vapour line) meets with each other is called as critical point. The triple point on P-V diagram is a line where all the phases, i.e. solid liquid and gas phases exist in equilibrium.

Fig.: P-V diagram of pure substance

At the pressure below than triple point substance can exist in liquid form. The

5 PROPERTIES OF PURE SUBSTANCE & GAS MIXTURE


region below the triple point line is solid vapour mixture region. For Water Critical pressure (Pc) = 221.2 bar Critical Temperature (Tc) = 374.150C 5.3 TRIPLE POINT Triple point is state at which solid, liquid and gas phases exist in equilibrium. On a P-T diagram, this condition is a point. But on P-V Diagram it is a line. The curve which shows the phase change of solid into liquid or liquid into solid is called as fusion curve. The curve which shows the phase change of liquid into vapour or vapour into liquid is called as vaporization curve.

Fig.: Triple point on P- T diagram The curve which shows the phase change of solid into vapour directly or vapour into solid is called as sublimation curve. The slop of vaporization and sublimation curve is positive. The slop of fusion curve for most of the substance is positive but for water is negative. For water, Triple point temperature (Tt) = 273.16 K Triple point pressure (Pt) = 0.6113 KPa 5.4 GIBBS PHASE RULE Gibbs phase rule describes the number of degrees of freedom, or the number of variables that must be fixed to specify the composition of a phase.

According to Gibbs phase rule, 2+C=F+P Where C = Number of chemically independent components F = Number of degrees of freedom of intensive properties P = Number of phases presented For a pure substance existing in a phase C = 1, P = 1 so, F = C + 2 – P = 1+2-1 = 2 It means there are two properties required to b3 known to fix up the state of the system at equilibrium. For a single component and two phase system (i.e. critical point), C =1, P = 2 F = F = C + 2 – P = 1+2-2 = 1 It means only one property is required to fix up the state of the system at equilibrium. if all three exists in equilibrium (i.e. triple point) then C = 1, P = 3 F = C + 2 – P = 1+2-3 = 0 This state is unique for a substance. It means at triple point all the intensive properties are fixed. 5.5 PHASE CHANGE OF PURE SUBSTANCE Thermodynamics deals only with liquid to gases to generate power. So it is important to understand the conversion of liquid into vapour or vice versa. Consider water at room temperature (20°C) and normal atmospheric pressure as shown in fig.

Fig. formation of steam The water is in liquid phase and it is called compressed liquid or sub cooled liquid.


Now heat is supplied to water, its temperature will increase. Due to the increase in temperature, the specific volume v, specific enthalpy h and specific entropy s will increase. As a consequence, the piston will move slightly upward therefore maintaining constant pressure. Now if continue heat is added to the water, the temperature will increase further until 100°C. At this point, any additional addition of heat will not increase the temperature of water; it will utilise to vaporize some water. This specific point where water starts to vaporize is called saturated liquid point. And the condition of the liquid is called as saturated liquid. If we continue to add heat to water, more and more vapour will be created, while the temperature and the pressure remain constant (T=100°C and P=1 atm). These conditions will remain the same until the last drop of liquid is vaporized. At this point, the entire cylinder is filled with vapour at 100°C. This state is called saturated vapour. The state between saturated liquid and saturated vapour where two phases exist is called saturated liquid-vapour mixture or wet vapour. After the saturated vapour phase, any addition of heat will increase the temperature of the vapour; this state is called superheated vapour. This concept can be applied to pure substance other than water. 5.6 T-S DIAGRAM OF PURE SUBSTANCE During the phase change of a pure substance as discussed in the previous topic first addition of heat is utilized to increase the temperature end entropy of the system. And then further addition of heat is to change the phase of liquid in which only entropy change temperature remains constant. But further addition of heat is to superheat the steam in which temperature and entropy both change. At a

given pressure process of phase change is shown T-S diagram.

Fig.:T-S diagram representing phase change for water at constant pressure Let us consider the different pressures and repeat process of phase change. On the different pressure we get the different saturated liquid point and saturated vapour point. The locus of all saturated liquid points is called as saturated liquid curve. And the locus of all the saturated vapour points at different pressure is called as saturated vapour curve. Where the saturated liquid and vapour curve meet with each other is called as Critical Point. The same process is shown on T-s and P-s diagram.

Fig. T-s diagram of pure substance


Fig. P-v diagram of pure substance 5.7. PROPERTIES OF PURE SUBSTANCE Saturation Temperature The temperature at which water starts boiling depends on the pressure. In other words, water starts boiling at 100 ºC but only at 1 atm. At different pressures, water boils at different temperatures. Saturation Pressure At a given pressure, the temperature at which a pure substance changes phase is called the saturation temperature (Tsat). Likewise, at a given temperature, the pressure at which a pure substance changes phase is called the saturation pressure (Psat). Sensible Heating

The addition of heat which is utilised to increase the temperature is called as sensible heating. In T- S diagram, Process

1-2 represents the sensible heat of water at 1 atmospheric pressure. At 1000C saturation point is attained. Latent Heating The addition of heat which is utilized to phase change is called as latent heating. In process 2-4, heat is added to water that is utilised to convert the phase of liquid in to vapour and it is called as latent heating. the latent heat decreases as pressure increases and at the critical point latent heat of vaporization is zero. Critical Point It is point at which liquid directly converts into vapour phase. At the critical point, latent heat is zero. For water at the critical point, Pcr = 220.8 bar Tcr = 374.140C Degree of Sub cooling If the actual temperature of liquid is less than the saturated temperature of liquid at given pressure the liquid is called as sub cooled liquid. The difference of saturated temperature and actual temperature of liquid is called as degree of sub cooling. Let the saturated temperature is Tsat. and actual temperature is Tsub.,then Degree of sub cooling = Tsat -Tsub

Degree of Sub cooling If the actual temperature of vapour is greater than the saturated temperature of vapour at given pressure the vapour is called as superheated vapour. The difference of actual temperature of vapour and saturated temperature is called as degree of superheating. Let the saturated temperature is Tsat. and actual temperature is Tsup., then Degree of superheating =Tsup - Tsat


Wet Vapour

Wet vapour is mixture of liquid and vapour phase. in wet vapour region liquid and vapour phase exists in equilibrium.

Dryness Fraction or quality of mixture (x)

If vapour is made wet by liquid droplets in suspension, it is important to know the degree of wetness. Any mass of wet consists of dry saturated liquid and dry saturated vapour. The ratio of mass of dry saturated vapour to total mass of water vapour is known as dryness fraction (x). It is also known as quality of steam. Let the mass of dry saturated vapour is mv and the mass of dry saturated liquid in wet vapour is lm , then dryness fraction

v

l v

mx

m m

Important points At saturated liquid line mass of vapour

is zero so value of x is zero along saturated liquid line.

At saturated vapour line mass of liquid is zero, so dryness fraction is 1. So dryness fraction varies from 0 to 1.

All dryness fraction lines converge at original point.

5.8 SPECIFIC VOLUME, ENTHALPY AND ENTROPY OF DIFFERENT PHASES The specific enthalpy entropy and specific volume are very important parameters to define the performance of any power producing system such as steam power plant working on Rankin cycle and power absorbing system such as vapour

compression refrigeration system. In most of the thermodynamic problem, these important phases can be existed. So it is important to determine the parameters in different region of pure substance.

1) Specific Volume, enthalpy and entropy of saturated liquid

In the T-S and P-h diagram it is shown by point 1. Enthalpy of this point is the heat supplied to increase the temperature from 00 C to saturated temperature at given constant pressure. these all parameters can be found Corresponding to saturated temp. at given pressure with help of stem table. Consider specific enthalpy, entropy and volume of saturated liquid at a given pressure and temperature are hf, sf and vf. specific enthalpy, entropy and volume of saturated vapour at a given pressure and temperature are hg, sg and vg. Then latent heat of vaporization hfg = hg - hf

specific volume Vfg = Vg - vf , Specific entropy Sfg = Sg - Sf From steam table at given pressure Enthalpy of saturated liquid h1 = hf Entropy of saturated liquid s1 = sf Volume of the saturated liquid v1 = vf

Fig. T- S diagram

Fig. P- h diagram


2) Specific Volume, enthalpy & entropy of sub cooled liquid

In the T-S and P-h diagram it is shown by point 2. Enthalpy of subcooled liquid

2 f p 1 1h h c T T ,

Entropy of subcooled liquid

15 f p n

2

TS S c l

T

Volume of subcooled liquid can be found at particular temperature at the given pressure from the steam table. 3) Specific Volume, enthalpy and entropy of saturated vapour

In the T-S and P-h diagram it is shown by point 4. From steam table at given pressure Enthalpy of saturated vapour

4 gh h ,

Entropy of saturated vapour 4 gs s

Volume of saturated vapour v4 = vg 4) Specific Volume, enthalpy and entropy of wet vapour

In the T-S and P-h diagram it is shown by point 3. Enthalpy of wet vapour

3 f fgh h x.h ,

Entropy of wet vapour

3 f fg f

LHs s xs s x

T

Volume of wet vapour v3= x.vg

5) Specific Volume, enthalpy and entropy of superheated vapour

In the T-S and P-h diagram it is shown by point 5. Enthalpy of superheated vapour

5 g p 5 4h h c T T

Entropy of superheated vapour

55 g p n

4

TS S c l

T

Volume of superheated vapour

v3= vg. 5

4

T

T

PROPERTIES OF GAS MIXTURE

5.9 EQUATION OF STATE

The ideal gas equation of state is P.v = RT can be established for the real gas with two important assumptions of the ideal gas. The ideal gas assumptions are: a) There is no attraction between the

molecules of the gas. b) The volume occupied by the molecules

is negligible. When the pressure is very low and temperature very large, the intermolecular attraction and volume of molecules has no more significance. As the pressure of gas increases, the intermolecular force forces of attraction and repulsion increases and also volume of the molecules becomes appreciable compare to total volume of the gas.

5.9.1 VANDER WAALS EQUATION

Vander Waals, by applying the law of mechanism of individual molecule introduced two correction terms in ideal gas equation. This is valid for the real gases.

2P (v b) RT

v

Where a and b are Vander Waals constant. a is introduced to account the existence of mutual attraction between molecules. The term a/v2 is called as force due to cohesion. The term b is introduced to account the volume of molecules is known as co – volume. 5.9.2 PROPERTIES OF CRITICAL POINT According to Vander Waals equation,

2

aP+ v-b =RT

v

or


2

RT aP

(v b) v

-------- (1)

Slop of P-v diagram at critical point is zero, so differentiating the equation with respect to v and putting equal to zero.

2 3

c

dp RT 2a0

dv vv b

2 3

cc

RT 2a

vv b

-------------- (II)

critical point is not only zero slop curve but at this point slop change so second derivative of P with respect to v is also zero. Differentiating the equation II and putting equal to Zero,

2

32 4

cc c

d p 2RT 6a0

dv vv b

3 4

cc

RT 3a

vv b

----------------- (III)

Now dividing the equation II by equation III, we get

c c

2v b v

3

Vc = 3b Substituting the value of vc in equation III, we get

2

3

2a 2b 8aRT

27b 27b

8aTc

27bR

Substituting the value of Tc and vc in equation I, we get

2

8aRX

a27bRP2b 9b

c 2

ap

27b

5.9.3 COMPRESSIBILITY FACTOR

The ideal gas equation can be given as Pv = RT If the gas is compressible (real gas), then

pv

ZRT

Z > 1 or Z < 1 5.9.4 COMPRESSIBILITY FACTOR AT CRITICAL POINT

At the critical point the compressibility factor is given by

c cc

c

P VZ

RT

Substituting the value of P, V and T at critical point, we get

2

c

aX3b

327bZ8a 8

R X27b R

c

3Z .375

8

5.10 PROPERTIES OF GAS MIXTURE

5.10.1 DALTON’S LAW OF PARTIAL PRESSURES

Let consider a homogeneous mixture of inert ideal gas at pressure P, Volume V and temperature T. Let us suppose there are n1

numbers of molecules of gas A, n2 number of molecules of gas B and n3 number of molecules of gas C.

fig. gas mixture


Then the equation of state for mixture of the gas is PV = (n1+n2+n3) ṜT Ṝ = universal gas constant = 8.314 KJ/ kg mol K Partial pressure of gas A, B and C can be written as

31 21 2 3

n .Tn .T n .TP ,P

RR R= ,P =

V V=

V

According to Dalton law of partial pressure

1 2 3P PP P=

31 2RR R n .Tn .T n .T

+ +V V

P=V

or KP=n R.T

V

where,

1k 2 3n n n n n

5.10.2 MOLE FRACTION The ratio of number of mole of a given gas to the total number of moles in the given mixture of ideal gas is known as mole fraction (x).

x1 = 1n

n , x2 = 2n

n and , x3 = 3n

n

Equivalent Gas constant (Re) Ideal gas equation for the gas mixture in the term of masses can be written as For gas A in the gas mixture P1V = m1R1T For gas B in the gas mixture P2V = m2R2T For gas C in the gas mixture P3V = m3R3T Using Dalton law of partial pressure, PV = P1V+ P2V+ P3V or PV = m1R1T + m2R2T + m3R3T PV = (m1R1 + m2R2 + m3R3)T-----------(I) Ideal gas equation for a gas mixture, PV = (m1 + m2 + m3).Re.T------------(II) Comparing the equation I and II, we get

1 1 2 2 3 3

1 2 3

m R m R m RRe

m m m

In the same way internal energy, enthalpy and specific heats are given as Specific Internal Energy of the mixture

1 1 2 2 3 3

1 2 3

m u m u m uUe

m m m

Specific enthalpy of the mixture

1 1 2 2 3 3

1 2 3

m h m h m hHe

m m m

Specific heat at constant pressure of the mixture

1 1 2 2 3 3

1 2 3

m Cp m Cp m CpCp

m m m

Specific heat at constant volume of the mixture

1 v1 2 v2 3 v3ve

1 2 3

m C m C m CC

m m m

Examples Q.1 Determine the specific enthalpy of

steam at 2 MPa and with a temperature of 2500C.

Solution: From the steam table

At 2 MPa , 0

satT 212.4 C

It must be the condition of super heating because the temperature of steam is higher than saturated steam. So Degree of superheating

sup satT -T =250-212.4=37.6K

Specific enthalpy of steam at 2MPa is given

gh 2797KJ / kg

So enthalpy

g p sup sath h c T T

= 2797.2 + 2.09 × 37.5 = 2875.9 KJ/kg


Q.2 Steam 0.95 dry at a pressure of 0.7 MPa is supplied to a heater through a pipe of 25mm internal diameter. The velocity in the pipe is 12 m/sec. water enters the heater at 190C , the steam is blow into it and the mixture of water and condensate leaves the heater at 900C. Calculate: a) mass of steam entering the heater b) mass of water entering the heater Properties of steam are:

Solution:

Specific volume of dry (0.95) at pr. 0.7 MPa is

1 gV xv 0.95 0.273

= 0.259 m3/kg a) Steam volume passing/sec

2πd l

4

23π

25 10 124

Steam volume passing /

2

r

πh 0.025 12 3600

4

mass of steam entering / hr

2π

0.025 12 36004

0.259

= 81.9 kg/hr b) Specific enthalpy of steam

entering heater is

2 f fgh h xh 697.1 0.95 2064.9

= 697.1 + 1961.7 = 2658.8 KJ / kg From the steam table at 900C

fh 376.8KJ / kg

And at 190C

fh 79.8KJ / kg

So for the heater applying energy balance equation

Enthalpy gained by water = Enthalpy lost by steam

376.8 79.8 81.9(2658.8 376.8) m

81.9 2282

m297

629.28 kg .m / hr

Q.3 1.5 kg of steam originally at

pressure of 1 MPa and temperature 2250C is expended until the pressure becomes 0.28 MPa. The dryness fraction at the steam is 0.9. Determine the change of internal energy.

Solution: At 1 MPa and 2250C, from steam table

3

1 1h =2886KJ/kg,v =0.2198m /kg

So6

1 1 1 1 3

1X10u =h -p v =2886- X0.2198

10

= 2886 – 219.8 = 2566.2 KJ/kg

At 0.28 MPa and dryness fraction 0.9

2 f 2 fg2h h xh

= 551.4 + 0.9 X 2170.1 = 551.4 + 1953.09

= 2504.49 KJ/kg Volume

2 g2v xv 0.9 0.646

= 0.581 3m / kg

Internal energy 2 2 2 2u h p v

3 2504.49 – (0.2 10 0.5 )8 814

= 2504.49 – 162.8 = 2341.69 KJ/kg Hence change of internal energy

2 1 2 1U U m u u

1.5 2341.69 – 2566.2

= - 489.77 KJ Q.4 The dryness fraction of steam at a

pressure of 2.2 MPa is measured using throttling calorimeter. After throttling, the pressure in the calorimeter is 0.13 MPa and the temperature is 1120C. Determine


the dryness fraction of steam at 2.2 MPa.

Solution: - From the steam table

Property of Superheated steam

Enthalpy of this stem after throttling can be found with help of linear interpolation

2h at the temp. of 1120C can be

found by linear interpolation

sup sat

2 g 150 g

sup sat

T Th h h h

T T

=2685.2+(2774.6-2685.8) (112 106.68)

(150 106.68)

= 2685.2 + 10.9 = 2696.1 KJ/kg Now for the wet steam before steam throttling

1 f1 1 fg1h =h +x h

= 931 + x1 X 1870 KJ/kg For the throttling process

1 2h h

931+(x1×1870) =2696.1

1

1765.1x

1870

1x 0.94

Q.5 Steams at 1.4 MPa and dryness fraction 0.7 is throttled to 0.11 MPa.

Determine the dryness fraction of steam after throttling.

Solution: For the throttling process 1 2h h

f1 1 fg1 f 2 2 fg2h x h h x h

830.1+0.7×1957.7=928.8+X2+2250.8

2

1771.7x

2250.8

2x 0.787

The steam becomes drier in this case.

Q.6 a) Determine the volume occupied

by 1 kg of steam at a pr. Of 0.85 MPa and having a dryness fraction of 0.97 . b) This volume is expanded

adiabatically to a pressure of 0.17 MPa, law of expansion is PV1.3 = C. Determine

i) Final dryness fraction of the steam

ii) Change of internal energy of system during the expansion.

Solution: Properties of system from steam table At 0.85 MPa , vg = 0.2268 m3/kg a) So

3

1 1 g1v =x v =0.97X0.2268=0.22m /kg

b) (i) Final dryness fraction

1.3 1.3

1 1 2 2PV P V

1

1.13

2

0.85V 0.22

0.17

2V 0.22 4.15

= 0.913 m3/kg At 0.17 MPa , vg2 = 1.031 m3/kg The steam is wet, so dryness fraction is

22

g2

u 0.913x 0.886

u 1.031

(ii) For the adiabatic expansion δQ = 0

Pr MPa

Sat.temp (0C)

Specific enthalpy kJ/kg

hg At 1500C

0.1 99.6 2675 2777

0.5 111.4 2693 2773


So change of internal energy

2 1U WU

1 1 2 2

2 1

P V P VU U

n 1

30.85 0.22 0.17 0.913 10

1.13 1

246 KJ / kg

There is a loss of internal energy.

Q.7 A mixture of ideal gases consisting

of 3 kg of nitrogen and 5 kg of carbon monoxide at a pressure of 300 KPa and temperature of 200C . Find (a) the mole fraction of constituent (b) Equivalent molecular weight of the mixture (c) Equivalent gas constant of the mixture, (d) the specific heat at constant pressure and volume of the mixture. Take 2γ 1.286forCO and

2γ 1.4forN .

Solution: let the gas 2CO as the gas 1 and gas

2N as the gas 2 is mixture.

The mole fraction for 2CO

1 11 1

1

x anM

d nn m

n

1

1

1 2

1 2

1x

m 5M 44 0.52

m m 5 3

44 28M M

Similarly the mole fraction for 2N

2

2

1 2

1 2

1x

m 3

M 28 0.48m m 5 3

44 28M M

(b) Equivalent molar weight of the mixture Me = x1M1 + x2M2 = 0.52×44 +0.48 × 28 = 36.32 Kg/Kg mole

(c).Equivalent gas constant of the mixture gas constant of gas 1

1

1

8.314

M 44

RR 0.19 KJ / Kg.K

2

1

RR 0.3

8.0

314

M 2KJ / K .K

8g

1 1 2 2

1 2

m R m R 44 0.19 28 0.30

m m 44Re 0.22 KJ /

28Kg.K

(d) The Equivalent specific heat at

constant pressure

1v1

R 0.190

γ 1 1.2C .66

86KJ K

1/ g.K

1P1 VC 1.286 6C γ 0.6

= 0.85 KJ/Kg.K

2V2

R 0.30

γ 1 1C 0.75

.4KJ / K

1g.K

VP2 2C 1.4 0. C 5γ 7

= 1.05 KJ/Kg.K The equivalent specific heat of mixture at constant pressure

1 p1 2 p2

p

2

e

1

m C m CC

m m

44 0.85 28 1.05

44 0.92

28 KJ / Kg

8.K

The equivalent specific heat of mixture at constant volume

1 V1 2 V2v

2

e

1

m C m CC

m m

44 0.66 28 0.75

44 28

= 0.695 KJ/Kg.K Q.8 from an experimental determination

the specific heat ratio for acetylene C2H2 is found to 1.26. Find the two specific heats. If heat is supplied to increase the temperature from 300 C to 600 C at constant pressure then find the increase in the enthalpy and entropy of the system.

Solution: Gas constant of acetylene

2 2C H (R)8.3143

kJ / kgKM 26

R


= 0.3198 kJ/kgK As adiabatic index (γ) = 1.26, then

We know that p

γc R

γ 1

1.26 KJ

0.3198 1.55 K1.26 1 Kg

And v

Rc

γ 1

0.3196

1.23KJ / kgK1.26 1

Change in enthalpy is given by

2 1 p 2 1h h c T T

KJ

30 4.1Kg

55 65.

Change in enthalpy is given by

2 p1

T2 P2c ln R ln

T1 1S S

P

p

T2c ln 0

T1

p

T2C ln

T1

333

ln303

1.55

= 0.146 KJ/Kg.K Q.9 One kg of air in a closed system,

initially at 50C and occupying 0.3 m3 volume, undergoes a constant pressure heating process to 1000C. There is no work other than Pdv work. Find (a) the work done during the process, (b) the heat transferred, and (c) the entropy change of the gas.

Solutions: T1 = 278 K V1 = 0.3 m3 M = 1 kg P1 = 265.95 kPa T2 = 1000C = 373 K P2 = 265.95 kPa

322

2

mRTV 0.40252m

P

(a) Work during the process

2

12 2 1

1

W pdV p V V 27.266kJ

(b) Heat transferred

1 2 2 1 12Q u u W

v 2 1 1 2mc T T W 95.476kJ

(c) Entropy change of the gas

2 22 1 p v

1 1

V pS S mc In mc In

V p

2p

1

V kJmc In 0.29543 .K

V kg


6.1 INTRODUCTION

In generally, refrigeration is defined as any process of heat removal. Refrigeration is defined as branch of Science that deals with the process of reducing and maintaining temperature of a space or material below the temperature of the surrounding. We see in nature that the heat spontaneously flows of heat from a high temperature body to a low temperature body. The reverse process to complete the thermodynamic cycle, in which heat Q will flow back from low temperature body to high temperature body, is not possible. So the logical conduction is that there must be a process in which some work is done.

6.1.1 NEED OF THERMAL INSULATION

Since heat well always flow from a region of high temperature to a region of lower temperature. There is always a continuous flow of heat into the refrigerated region from the warmer surroundings. To limit the heat flow into refrigerated region, it is usually necessary to isolate the region from its surrounding with a good insulation.

6.1.2 REFRIGERATION LOAD

The rate at which heat must be removed from refrigerated space or material in order to produce and maintain the desired temperature condition is called as refrigeration load or cooling load. The total refrigeration load is the sum of heat gains from different sources:- (1) Heat transmitted by conduction

through the insulated walls. (2) Heat that must be removed from warm

air that enters the space through openings and closing doors.

(3) Heat that must be removed from refrigerated product to reduce the temperature of products to the storage temperature.

6.1.3 REFRIGERANT

In the refrigeration process, the substance employed as the heat absorber or coaling agent is called the refrigerant. When the absorbed heat causes an increase in temperature of refrigerant, cooling process is said to be sensible, whereas when the absorbed heat causes a change in physical state of refrigerant, cooling effect is said to be latent. Ice has been used successfully for many years as a refrigerant. Not to many years ago ice was the only cooling agent available for use in domestic and small commercial refrigerator.

6.1.4 UNIT OF REFRIGERATION

It is given by Tonnes of Refrigeration (TR). 1TR is defined as refrigeration effect produced by uniform melting of 1000 kg of ice from and at 00C in 24 hours. 1 TR = 1000×334 KJ refrigeration effect in

1000 334

24 36024 hours 3.86

0KJ / Sec.

For practical problem it is taken as 3.5KJ/Sec or 210 KJ/Minute.

6.1.5 CO EFFICIENT OF PERFORMANCE

The co efficient of performance (COP) of refrigerator is ratio of heat extracted in the refrigerator to the work done on the refrigerant working for the refrigeration cycle.

Heat extracting from refrigerated space Q

Work DoCOP

ne W

6.2 AIR STANDARD REFRIGERATION CYCLE

Air cycle refrigeration systems belong to the general class of gas cycle refrigeration systems, in which a gas is used as the working fluid. The gas does not undergo

6 REFRIGERATION


any phase change during the cycle; consequently, all the internal heat transfer processes are sensible heat transfer processes. Gas cycle refrigeration systems find

applications in air craft cabin cooling and It also in the liquefaction of various gases. 6.2.1 AIR STANDARD CYCLE ANALYSIS Air cycle refrigeration system analysis is considerably simplified if one makes the following assumptions: i. The working fluid is a fixed mass of air

that behaves as an ideal gas ii. All the processes within the cycle are

reversible, i.e., the cycle is internally reversible.

iv. The specific heat of air remains constant throughout the cycle.

First of all we study the ideal refrigeration cycle i.e. Reversed Carnot Cycle. 6.2.2 A REFRIGERATOR WORKING ON REVERSED CARNOT CYCLE: Reversing the Carnot cycle does reverse the directions of heat and work interactions. A refrigerator or heat pump that operates on the reversed Carnot cycle is called a Carnot refrigerator or a Carnot heat pump. Reversed Carnot cycle is shown in Fig. It consists of the following processes.

Fig. a refrigerator working on reversed Carnot Cycle Process 4-1: Absorption of heat by the working fluid from refrigerator at constant

low temperature TL during isothermal expansion. Process 1-2: Isentropic compression of the working fluid with the aid of external work. The temperature of the fluid rises from TL to TH. Process 2-3: Isothermal compression of the working fluid during which heat is rejected at constant high temperature TH. Process 3-4: Isentropic expansion of the working fluid. The temperature of the working fluid falls from TH to TL.

Fig. P-V and T-S Diagram of Reversed Carnot Cycle The reversed Carnot cycle is the most

efficient refrigeration cycle operating between two specified temperature levels. It sets the highest theoretical COP. The coefficient of performance for Carnot refrigerators and heat pumps are:

COP of Refrigerator

R

Heat absorbed

Work supplCOP

ied

Heat absorbed

Heat rejected Heat absorbed

L

L

H Q

Q

Q

L

L

H(dS) T (dS)

T (dS)

T

So,

L LR

H L H L

Q TCOP

Q Q T T

-----(7.1)


COP of Heat Pump

HP

Heat Rejected

Work suppCOP

lied

Heat Rejected

Heat rejected Heat absorbed

L

H

H Q

Q

Q

L

H

H(dS) T (dS)

T (dS)

T

So H HHP

H L H L

Q TCOP

Q Q T T

--(7.2)

From the equation 7.1 and 7.2

HP RCOP COP 1 ------------- (7.3)

Notice that turbine is used for expansion process between higher and lower temperature limit. Work done by the turbine helps supply some of the work required by the compressor from the external source.

Practically, the reversed Carnot cycle cannot be used for refrigeration purpose as the isentropic process requires very high speed operation, whereas the isothermal process requires very low speed operation.

6.3 REFRIGERATOR WORKING ON REVERSED BRAY TON CYCLE

This is an important cycle frequently employed in gas cycle refrigeration systems. This may be thought of as a modification of reversed Carnot cycle, as the two isothermal processes of Carnot cycle are replaced by two isobaric heat transfer processes. This cycle is also called as Joule or Bell-Coleman cycle. Figure shows the schematic of a closed reverse Bray ton cycle and also the cycle on T-s diagram.

Fig. Reversed Bray ton Refrigeration System

There are four important processes in reversed Bray ton Cycle: Process 1-2: Reversible adiabatic compression in a compressor Process 2-3: Reversible isobaric heat rejection in a heat exchanger Process 3-4: Reversible adiabatic expansion in a turbine Process 4-1: Reversible isobaric heat absorption in a heat exchanger All the Processes are on P-V and T-S diagram.

Fig. Reversed Bray ton cycle on T-S diagram Process 1-2: Gas at low pressure is compressed isentropic ally from state 1 to state 2. Applying steady flow energy equation and neglecting changes in kinetic and potential energy, W1-2 = m (h2-h1) = m. Cp. (T2-T1) But process 1-2 is reversible adiabatic Process or isentropic process. So,

γ 1γ 1γ

2 2 γp

1 1

T Pr

T( 4)

P7.

Where rp

= (P2/P

1) = pressure ratio

Process 2-3: Hot and high pressure gas flows through a heat exchanger and rejects heat sensibly and isobaric to a heat sink. The enthalpy and temperature of the gas drop during the process due to heat exchange, no work transfer takes place and the entropy of the gas decreases. Q2-3 = m (h2-h3) = m. Cp. (T2-T3) Process 3-4: High pressure gas from the heat exchanger flows through a turbine undergoes isentropic expansion and


delivers net work output. The temperature of the gas drops during the process from T

3

to T4. From steady flow energy equation:

W3-4 = m (h3-h4) = m. Cp. (T3-T4) But process 3-4 is reversible adiabatic Process or isentropic process. So,

γ 1γ 1γ

3 3 γp

4 4

T Pr

T P

------------- (7.5)

Where rp

= (P3/P

4) = pressure ratio

So, 3 2

4 1

T T

T T

Process 4-1: Cold and low pressure gas from turbine flows through the low temperature heat exchanger and extracts heat sensibly and isobaric ally from a heat source, providing a useful refrigeration effect. The enthalpy and temperature of the gas rise during the process due to heat exchange, no work transfer takes place and the entropy of the gas increases. The refrigerating effect by the refrigerant can be given as: 6.3.1 REFRIGERATING EFFECT

Q4-1 = m (h1-h4) = m. Cp. (T1-T4) COP of reversed Bray ton cycle can be given:

COP Refrigerating Effect / NetWork Done

4 1

1 2 3 4

Q

W W

p. 1 4

p. 2 1 p. 3 4

m.c (T T )

m.c (T T ) m.c (T T )

1 4

2 3 1 4

(T T )

(T T ) (T T )

Solving the equation

41

1

3 42 1

2 1

TT 1

T

T TT 1 T 1

T T

1

γ 122 1 γ

p1

T 1 1

TT T1 r 1

T

So,

p

1COP

1r 1

From the above expression for COP, the following observations can be made:

Fig. Comparison between Reversed Bray ton Cycle and Carnot Cycle.

For fixed heat rejection temperature (T

3) and fixed refrigeration temperature

(T1), the COP of reverse Bray ton cycle

is always lower than the COP of reverse Carnot cycle as shown in fig.

COP of Bray ton cycle approaches COP of Carnot cycle as T

1 approaches T

4

(thin cycle), however, the specific refrigeration effect [c

p(T

1-T

4)] also

reduces simultaneously. COP of reverse Bray ton cycle decreases

as the pressure ratio rp

increases.

6.3.2 ACTUAL REVERSE BRAY TON CYCLE

The actual reverse Bray ton cycle differs from the ideal cycle due to: Non-isentropic compression and

expansion processes. Pressure drops in cold & hot heat exchangers.

Fig. Actual Reversed Bray ton Cycle


Figure shows the ideal and actual cycles on T-s diagram. Due to this irreversibility, the compressor work input increases and turbine work output reduces. If η

compressor ,isen and η

turbine ,isen are the

isentropic efficiencies of compressor and turbine, respectively, these are defined as:

2 1 2 1

' '

2

Copm

1 2 1

.

h h T T

h h T T

3 4 3 4

' '

3

Turb

4 4

i e

3

n

h h T

T

T

h h T

Thus the net work input increases due to increase in compressor work input and reduction in turbine work output. The refrigeration effect also reduces due to the irreversibility’s. As a result, the COP of actual reverse Bray ton cycles will be considerably lower than the ideal cycles. Design of efficient compressors and turbines plays a major role in improving the COP of the system. In practice, reverse Bray ton cycles can be open or closed.

6.3.3 AIRCRAFT COOLING SYSTEMS

An aircraft, cooling systems are required to keep the cabin temperatures at a comfortable level. Even though the outside temperatures are very low at high altitudes, still cooling of cabin is required due to: Large internal heat generation due to

occupants, equipment etc. Heat generation due to skin friction

caused by the fast moving aircraft At high altitudes, the outside pressure

will be sub-atmospheric. When air at this low pressure is compressed and supplied to the cabin at pressures close to atmospheric, the temperature increases significantly.

Solar radiation 6.3.4SIMPLE AIRCRAFT REFRIGERATION CYCLE Figure shows the schematic of a simple aircraft refrigeration system and the

operating cycle on T-S diagram. This is an open system. As shown in the T-S diagram, the outside low pressure and low temperature air is compressed due to ram effect to ram pressure in the process 1-2. During this process its temperature increases from T1 to T2. This air is compressed in the main compressor in process 2-3, and is cooled to temperature T4 in the air cooler. Its pressure is reduced to cabin pressure in the turbine, as a result its temperature drops from T4 to T5. The cold air at stateT5 is supplied to the cabin. It picks up heat as it flows through the cabin providing useful cooling effect. The power output of the turbine is used to drive the fan, which maintains the required air flow over the air cooler. This simple system is good for ground

cooling.

fig: Simple air craft refrigeration system

'22

1

T γ 11 M

T 2

Where M is the Mach number, which is the ratio of velocity of the aircraft (V) to the sonic velocity C

Mach number: 1

V VM

C RT

6.4VAPOUR COMPRESSION REFRIGERATION CYCLE

Vapour compression cycle is an improved type of air refrigeration cycle in which a suitable working substance, termed as refrigerant, is used. The refrigerant used, does not leave the system, but is circulated throughout the system alternately condensing and evaporating.


In evaporating, the refrigerant absorbs its latent heat from the solution which is used for circulating it around the cold chamber. In condensing, it gives out its latent heat to the circulating water of the cooler. Vapour compression refrigeration system is now-a-days used for all purpose refrigeration. It is used for all industrial purposes from a small domestic refrigerator to a big air conditioning plant. 6.4.1 SIMPLE VAPOUR COMPRESSION REFRIGERATION SYSTEM It consists of the following essential parts:

Fig. Simple vapour compression refrigeration system Compressor The low pressure and temperature vapour refrigerant from evaporator is drawn into the compressor through the suction valve A, where it is compressed to a high pressure and temperature. This high pressure & temperature vapour refrigerant is discharged into the condenser through the discharge valve B. Condenser The condenser consists of coils of pipe in which the high pressure and temperature vapour refrigerant is cooled and condensed. The refrigerant, while passing through the condenser, gives up its latent

heat to the surrounding condensing medium which is normally air or water. Receiver The condensed liquid refrigerant from the condenser is stored in a vessel known as receiver from where it is supplied to the evaporator through the expansion valve or refrigerant control valve.

Expansion Valve

It is also called throttle valve or refrigerant control valve. The function of the expansion valve is to allow the liquid refrigerant under high pressure and temperature to pass at a controlled rate after reducing its pressure and temperature. Some of the liquid refrigerant evaporates as it passes through the expansion valve, but the greater portion is vaporized in the evaporator at the low pressure & temperature

Evaporator An evaporator consists of coils of pipe in which the liquid-vapour. refrigerant at low pressure and temperature is evaporated and changed into vapour refrigerant. In evaporating, the liquid vapour refrigerant absorbs its latent heat of vaporization from the medium which is to be cooled.

6.4.2 THEORETICAL VAPOUR COMPRESSION CYCLE

A vapour compression cycle with dry saturated vapour after compression is shown on T-S and P-h diagrams as shown in Figures. At point 1, let the temperature, pressure and entropy of the vapour refrigerant is T1, p1 and S1 respectively. The four processes of the cycle are as follows:


Fig: VCR cycle on T-s and P-h diagram Process 1-2-Compression Process The vapour refrigerant at low pressure P1 and temperature T1 is compressed isentropic ally to dry saturated vapour as shown by the vertical line 1-2 on the T-s diagram and by the curve 1-2 on p-h diagram. The pressure and temperature rise from P1 to P2 and T1 to T2 respectively. The work done during isentropic compression per kg of refrigerant is given by w = h2 – h1 Where h1 = Enthalpy of vapour refrigerant at temperature T1, i.e. at suction of the compressor, and h2=Enthalpy of the vapour refrigerant at temperature T2. i.e. at discharge of compressor. Process 2-3- Condensing Process The high pressure and temperature vapour refrigerant from the compressor is passed through the condenser where it is completely condensed at constant pressure P2 and temperature T2 as shown by the horizontal line 2-3 on T-s and p-h diagrams. The vapour refrigerant is changed into liquid refrigerant. The refrigerant, while passing through the condenser, gives its latent heat to the surrounding. Process 3-4 - Expansion Process The liquid refrigerant at pressure P3 = P2

and temperature T3 = T2, is expanded by throttling process (Isenthalpic Process) through the expansion valve to a low pressure P4 = P1 and Temperature T4 = T1 as shown by the curve 3-4 on T-s diagram and by the vertical line 3-4 on P-h diagram.

Some of the liquid refrigerant evaporates as it passes through the expansion valve, but the greater portion is vaporized in the evaporator. We know that during the throttling process, no heat is absorbed or rejected by the liquid refrigerant. Process 4-1-Vaporizing Process The liquid-vapour mixture of the refrigerant at pressure P4=P1 and temperature T4=T1 is evaporated and changed into vapour refrigerant at constant pressure and temperature, as shown by the horizontal line 4-1 on T-s and p-h diagrams. During evaporation, the liquid-vapour refrigerant absorbs its latent heat of vaporization. The heat which is absorbed by the refrigerant is called refrigerating effect and it is briefly written as RE. The refrigerating effect or the heat absorbed or extracted by the refrigerant during evaporation per kg of refrigerant is given by RE = h1 – h4 = h1 – hf3 Where hf3 = Sensible heat at temperature T3, i.e. enthalpy of liquid refrigerant leaving the condenser. CO Efficient of Performance Coefficient of performance, C.O.P. = (Refrigerating effect)/ (Work done)

1 f 31 4

2 1 2 1

h hh hO.P

h h h h

6.5 EFFECT OF PARAMETERS ON COP OF VAPOUR COMPRESSION REFRIGERATION CYCLE 1. Effect of Suction Pressure The suction or evaporator pressure decreases due to the frictional resistance of flow of the refrigerant. Let us consider a theoretical vapour compression cycle 1-2-3-4 when the suction pressure decreases from Ps to Ps’ as shown on p-h diagram in Figure. It may be noted that the decrease in suction pressure:


a) Decreases the refrigerating effect from (h1 – h4) to (h11 – h4). b) Increases the work required for

compression from (h2–h1) to (h21– h11).

fig. Effect of suction pressure

Since the C.O.P, of the system is the ratio of refrigerating effect to the work done, therefore with the decrease in suction pressure, the net effect is to decrease the C.O.P. of the refrigerating system for the same refrigerant flow. Hence with the decrease in suction pressure The refrigerating capacity of the system

decreases. The refrigeration cost increases.

2. Effect of Discharge Pressure In actual practice, the discharge or condenser pressure increases due to frictional resistance of flow of the refrigerant. Let us consider a theoretical vapour compression cycle l-2-3-4 when the discharge pressure increases from PD to PD

1 as shown on p-h diagram in Figure resulting in increased compressor work from (h2 – h1) to (h21 – h1) and decreasing refrigeration effect from (h1 – h4) to (h1 – h41). So COP is ratio of refrigeration effect to work done. So COP of the cycle decreases.


6.5.1 EFFECT OF LIQUID SUB COOLING It is possible to reduce the temperature of the liquid refrigerant to within a few degrees of the temperature of the water entering the condenser. In some condenser designs it is achieved by installing a sub-cooler between the condenser and the expansion valve. The effect of sub-cooling of the liquid from t3 = tk to t31 is shown in Figure. It will be seen that sub-cooling reduces flashing of the liquid during expansion and increases the refrigerating effect. But there is no effect on the work done by the compressor. And the COP of the cycle is ratio of refrigeration effect to work done by the compressor. So the COP of the cycle increases in sub-cooling of the liquid.


6.6VAPOURABSORPTION REFRIGERATION SYSTEMS The Refrigeration effect by Vapour Compression Refrigeration System is an efficient method. But the input energy given in the VCR system is work i.e. high grade energy and therefore very expensive. So In Vapour Absorption Refrigeration Systems, the mechanical work is replaced by the heat. Hence these systems are also called as heat operated or thermal energy driven systems. Since these systems run on low-grade thermal energy, they are preferred when low-grade energy such as waste heat or solar energy is available.


Since conventional absorption systems use natural refrigerants such as water or ammonia they are environment friendly.

6.6.1 WORKING OF SIMPLE VAPOUR ABSORPTION REFRIGERATION CYCLE The systematic diagram of VAR system is shown. In this, compressor is replaced by absorber system (i.e. Absorber, Pump, Heat Exchanger Heat generator and rectifier). Working of the system is almost same as the working of VCR system excepting the functionality of compressor. The working can be explained with help of sketch diagram. The low temperature refrigerant enters the evaporator and absorbs the required heat from the evaporator and leaves the evaporator as saturated vapour. Slightly low pressure NH3 vapour is absorbed by the weak solution of NH3 which is sprayed in the absorber as shown in fig.

Fig. Vapour Absorption Refrigeration System Weak NH3 solution (i.e. know as aqua–ammonia) enters in to absorber and becomes strong solution after absorbing NH3 vapour. Then it is pumped to the generator through the heat exchanger. The pump increases the pressure of the strong solution. The strong NH3 solution absorbs heat form high temperature weak NH3 solution in the

heat exchanger. The weak high temperature ammonia solution from the generator is passed to the heat exchanger through the pressure reducing valve. The pressure of the liquid is reduced to the absorber pressure by the throttle valve. Ammonia vapour is produced in the generator at high pressure by a heating generator. The water vapour carried with ammonia is removed in the rectifier and only the dehydrated ammonia gas enters into the condenser. High pressure NH3 vapour is condensed in the condenser. The cooled NH3 solution is passed through a throttle valve and the pressure and temperature of the refrigerant are reduced below the temperature to be maintained in the evaporator. The refrigerant absorbs the heat from the medium and converts in to the vapour phase and again it goes to generator. 6.6.2 COP OF IDEAL VAR CYCLE The working of VAR system is shown in fig. Let us consider the Heat QE is absorbed by the refrigerant in absorber at temperature of TE. In the heating generator, QG amount of heat is added at temperature TG to the refrigerant. The QC amount heat is rejected in condenser at temperature of TC. Pump work is WP. The COP of system is given as

VAR

Refrigeration Effect

Given COP

t

inpu

E E

G P G

Q Q

Q W Q

Considering a reversible refrigeration cycle, QC = QG + QE -----------(1) for a reversible cycle Clausius Inequality Law,

C GE

C E G

Q QQδQ0,

T T T T ∮

Substituting the value from equation 1

G GE

C E G

Q QE QQ

T T T


Or, G GE E

C E G C

Q QQ Q

T T T T

Or, E C E G G C

C G

Q (T T ) Q (T T )

T T

Or, G C CE

G G C E

(T T ) TQ

Q T (T T )

CarnotVAR CarnotCOP COP

6.7 REFRIGERANT Refrigerant is the fluid used for heat transfer in a refrigerating system that absorbs heat during evaporation from the region of low temperature, and releases heat during condensation at a region of higher temperature. Due to several environmental issues such as ozone layer depletion and global warming and their relation to the various refrigerants used, the selection of suitable refrigerant has become one of the most important issues in recent times. For Example, Air used in an air cycle refrigeration system can also be considered as a refrigerant. 6.7.1 CLASSIFICATION OF REFRIGERANT

There are two types of refrigerants 1. Primary Refrigerant 2. Secondary Refrigerant

1. Primary Refrigerant

Primary refrigerants are those fluids, which are used directly as working fluids, for example in vapour compression and vapour absorption refrigeration systems.

2. Secondary Refrigerant

Secondary refrigerants are those liquids, which are used for transporting thermal energy from one location to other. Secondary refrigerants are also known under the name brines or antifreezes. Generally, the freezing point of brine will be lower than the freezing point of its constituents. So these are used as Secondary refrigerants.

If the operating temperatures are above

0oC, then pure water can also be used as

secondary refrigerant, The commonly used secondary refrigerants are:

The solutions of water and ethylene glycol,

The solutions of propylene glycol or calcium chloride.

6.7.2 PROPERTIES OF REFRIGERANT There should be the following properties of the refrigerants: a) Latent heat of vaporization: it should be as large as possible so that the required mass flow rate per unit cooling capacity will be small.

b) Isentropic index of compression: It should be as small as possible so that the temperature rise during compression will be small.

c) Vapour specific heat: It should be large so that the degree of superheating will be small.

d) Thermal conductivity: Thermal conductivity in both liquid as well as vapour phase should be high for higher heat transfer coefficients.

e) Viscosity: Viscosity should be small in both liquid and vapour phases for smaller frictional pressure drops.

f) Ozone Depletion Potential (ODP): According to the Montreal protocol, the ODP of refrigerants should be zero, i.e., they should be non-ozone depleting substances. Refrigerants having non-zero ODP have either already been phased-out (e.g. R11, R 12).

g) Global Warming Potential (GWP): Refrigerants should have as low GWP value as possible to minimize the problem of global warming. h) Toxicity


The refrigerants used in a refrigeration system should be non-toxic. 6.8 DESIGNATION OF REFRIGERANTS There are following types of refrigerants. The designation of these refrigerants can be done by following manner: 1) Halocarbon compounds: These refrigerants are derivatives of alkenes (C

nH

2n+2) such as methane (CH

4),

ethane (C2H

6). These refrigerants are

designated by R-ABC Where: A+1 indicates the number of Carbon (C) atoms B-1 indicates number of Hydrogen (H) atoms, and Z indicates number of Fluorine (F) atoms Example: R- 22 A = 0 ⇒ No. of Carbon atoms = 0+1 = 1 B = 2 ⇒ No. of Hydrogen atoms = 2-1 = 1 C = 2 ⇒ No. of Fluorine atoms = 2 The balance = 4 – no. of (H+F) atoms = 4-1-2 = 1, So No. of Chlorine atoms = 1 The chemical formula of R 22 = CHClF

2 The commonly used refrigerant i.e. R-134a = C

2H

2F

4

2) Inorganic refrigerants: These are designated by number 7 followed by the molecular weight of the refrigerant. Examples: (a) Ammonia: Molecular weight is 17, the designation is R -717 (b) Carbon dioxide: Molecular weight is 44, the designation is R -744 (c) Water: Molecular weight is 18, the designation is R -718 3) Azeotropic Refrigerant: Azeotropic mixtures are designated by 500 series, these are the mixture of two or more halocarbon compound. Some Azeotropic mixtures:

R -500: Mixture of R 12 and R 152a R -502: Mixture of R 22 and R 115 R-503: Mixture of R 13 and R 23

(4) Hydrocarbons: Propane (C

3H

8) : R 290

n-butane (C4H

10) : R 600

Iso-butane (C4H

10) : R 600 a

Examples Q.1 In reversed Bray ton cycle, the

temperature at the end of heat absorption and heat rejection are 00C and 300C. The pressure ratio is 4 and the pressure in the cooler is 4 bar, determine the temperature at all the states and volume flow rates at inlet of compressor and at exit of turbine for 1 TR cooling capacity.

Solution: T1 = 273 K T3 = 303 K Temperature at all points

γ 1

2 γ

1

Tr

T

0.4

1.42T 273 4 405.67K

γ 1

3 γ

4

Tr

T

4 0.4

1.4

303T 203.9K

4

Volume flow rate of air for 1 TR or 3.5KJ/sec

Refrigeration effect a p 1 4m c T T

p3.53.5 3.5 m X1.005(273 203.9)

pm 0.05kg / sec

At the inlet of compressor density,

5

11

1

P 30P 1.27kg / sec.

RT 287X273


Volume flow rate V = P1 X ma = 1.27 X 0.05

= 0.064 m3/sec. Q.2 The compressed air from main

compressor of an air craft cooling system is bled off at 4.5 bar and 2000C. It is then passed through the heat exchange in which the ram air is forced through for cooling purpose by fan driver by cooling turbine. The condition of the inlet of cooling turbine is 4 bar an 300C. The air expands in cooling turbine is 4 bar an 300C. The air expands in cooling turbine up to 0.7 bar. The isentropic efficiency of cooling turbine is 80% and flow rate through cooling turbine is 30 kg/min. Find a) Actual exit temperature from

cooling turbine b) The power delivered to ram air

blow fan c) Tons of refrigeration, if

temperature of cabin cockpit area is 250C.

Solution:- 1) Exit temp. Of turbine

γ 1

γ 0.4

1.42 2

3

3 3

T P 0.7T 300 184.15K

T P 4

T3 = 184.15 K Efficiency of the turbine

1

2 3

2 3

T T0.8

T T

1

3 2 3 2T T T 0.8 T

1

3T 303 0.8(303 184.15)

1

3T 207.92k

b) Power delivered to fan = work output from turbine

1

2 3

301.005 T T

60

= 0.5 × 1.005 (303 – 207.92) = 47.78 KW c) Refrigeration effect

1

a p cabin 3RE m c T T

= 0.5 × 1.005 (298 – 207.92) =45.27 KJ/s =45.27/3.5 = 12.9 TR

Q.3 A dense air refrigeration machine operating on Bell-Coleman cycle operates between 3.4 bar and 17 bar. The temperature of air after the cooler is 150C and after the refrigerator is 60C. The refrigeration capacity is 6 Tonnes, calculate: 1) temperature after the

compression and expansion 2) air circulation per minute 3) work of compressor and

expander 4) Theoretical COP.

Solution:

0

3T 15 C 288K

0

1T 6 C 279K

RE = 6 tonnes (I) temp. After comp. & expansion

γ 1

γ 0.4

1.42 2

1 1

T P 17

T P 3.4

T2=279×1.5=445.69 k

V 1 0.4

V 1.44 4

3 3

T P 3.4.62

T P 17

T4 = 288× 0.62 =180.2 K

(II) Air circulation per minute

Refrigeration effect= a p 1 4m c T T

6 ×210 = ma ×1.005(279-180.2) Ma=12.7 kg/min. (III) Work of compressor:-


comp a 2 1

nw m R T T

n 1

1.4 12.7287(445.29 279)

1.4 1 60

= 35.44 KW Work of expander

turbine a 3 4

nw m R T T

n 1

1.4 12.7

287(288 180.2)1.4 1 60

= 22.89 KW (IV) Cop of cycle

Heatabsorbed 6X3.5

Copworkdone 35.44 22.89

COP = 1.67 Q.4 A simple saturation cycle using R-12

is desired taking a load of 10 Tons. The refrigerating and ambient temperature are 00C and 300C. A minimum temperature difference of 50C is required in evaporator and condenser for heat transfer. Find

a) Mass flow rate through system b)Power required in kw. The properties of R-12 are following:-

Sat. Temp.

0C

Sat. Pr.

bar

Specific volume m3/kg

Sat (10-3) Sat. vapour

-50C 2.61 0.71 0.0650

00C 3.08 0.72 0.0554

300C 7.45 0.77 0.0235

350C 8.47 0.79 0.0206

Enthalpy kJ/kg Entropy kJ/kg. k

Sat. Liquid

Sat. vapour

Sat. Liquid

Sat. vapour

31.4 185.4 0.1251 0.6991 36.1 187.5 0.1420 0.6966

64.4 199.6 0.2399 0.6854

69.5 201.5 0.2559 0.6839

Specific heat of vapour R-12 is 0.95 KJ/KgK Solution:-

Temp. Of refrigerant in evaporator is -50C and in the condenser is 350C.

At low pressure

1 gh h at -50C

1h 185.4kJ / kg

Enthalpy of point 4 4 3 f 3h h h 69.5kJ

Refrigerant effect

1 4m h h m 185.4 69.5

m 115.9kJ / sec 10 3.5 115.9 m

Mass flow rate 35

m 0.3kg / sec.115.9

In process(1-2)i.e.isentropic process 1 2s s

sap

g1 g2 p n

sat

Ts s c l

T

sap

p n

T0.6991 0.6839 c l

308

supT 312.9 k

h2 = hg2 + cp Tsap − Tsat

= 201.5 + 0.95 (312.9-308) = 206.25 KJ/Kg (II) Power required by compressor

2 1w m h h

0.3 (206.25 185.4) 6.3KW

Q.5 An ideal vapour compression

refrigerator using R-12 operates between temperature limits of -100C and 400C. The refrigerator leaves the condenser dry saturated. The rate of flow of refrigerant through the unit is 150 Kg/hr. Calculate the refrigerating effect per kg of refrigerant , COP If 1) refrigerant leaves the condenser

as dry saturated 2) refrigerant is sub cooled to 200C

before throttling.


Assume that enthalpy of refrigerant before throttling is approximately equal to enthalpy of refrigerant at under cooled temperature of 200C and it as hf = 50.59 kJ/kg.

Solution:

(1) if the refrigerant leaves dry saturated h1 = 183.19 kJ/kg

1s 7019kJ / kg.k

2 1s s

sap

1 g p n

sat

Ts s c l

T

At 400C Sg = 0.6825 Cp = 0.609 kJ/kgk

sap

1 n

Ts 0.6825 0.609l

313

sap

n

T0.7019 0.6825 0.609l

313

sapT 323.1K

2 g p sap sath h c T T

= 203.20+ 0.609 (323.1-313)

2h 209.37kJ / kg.k

3 f 3 4h h h 74.59kJ / kg

(i) Refrigerating effect = h1 – h4 = 183.1 – 74.59

= 108.6 KJ/kg

1 4

2 1

h h 108.6Cop 4.15

h h 209.6 183.1

(iii) off refrigerant is sub cooled

1

3 4h 50.59 h

Refrigerating effect = h1 – h4 = 183.1– 50.59

= 132.51 KJ/kg

1 4

2 1

h h 132.51Cop 5.06

h h 209.6 183.19

Q.6 A two cylinder R-12 compressor has

bore and stroke equal to 5.65 cm and 5 cm. Its rpm is 1450 rpm.If liquid refrigerant is throttled by throttle valve at 400C. Determine the mass of refrigerant circulated per minute and theoretical capacity when suction temperature is –100C. Assume γ = 1.13 , properties of R-12 are:

Temp Pr.(bar) Volume vapour m3/kg

Enthalpy liquid kJ/kg

-100C 2.1928 0.07702 190.72

400C 9.5944 0.01863 239.29

Enthalpy vapour kJ/kg

Entropy liquid kJ/kgk

Entropy vapour kJ/kgk

347.96 0.9656 1.5632

368.67 1.1324 1.5456

Solution: Theoretical piston displaced volume

2

P

πV d LNk

4

2

2 2π5.65X10 5 1450 2 10

4

= 0.335 m3/min. Vg = 0.07702 m3/kg at -100C

So refrigerant mass circulated/min at -100C suction temperature = 0.335/0.07702 = 4.6 kg/min h1 = 347.96 kJ/kg h4 = 239.29 kJ/kg

Refrigerating effect

1 4RE m h h

= 4.6 X (347.96 – 239.29) = 500.8 kJ/min Q.7 A R-12 refrigeration machine has

saturated suction temperature of


50C and saturated discharge temperature of 400C. Determine i) COP ii) Theoretical power. Per ton of

refrigeration when the compression is dry properties are:

Temp. Pr. bar Volume m3/kg

hf

50C 3.6375 0.048 204.68 400C 9.5909 0.0184 239.03

Enthalpy vapour kJ/kg

Entropy liquid kJ/kgk

Entropy vapour kJ/kgk

347.96 0.9656 1.5632 368.67 1.1324 1.5456

Solution: Process 1-2 is isentropic process.

1 2S S

sap

g2 p n

sat

T1.5569 S c l

T

sap

n

T1.5569 1.5569 0.762l

313

supT 317.5k

2 g2 p sap sath h c T T

= 368.87 + 0.762 (317.5 – 313) = 372.3 kJ/kg 3 fh h at 313K =239.03 kJ/kg = h4

Refrigerating effect 1 4m h h

1 4

210m h h

60

m 0.03kg / sec

Theoretical power per ton of refrigeration

2 1P m h h

0.03 372.3 354.885 0.53KW / TR

1 4

2 1

h h 354.885 239.03COP 6.65

h h 372.3 354.885

Q.8 Assume a simple saturated cycle with suction saturation temp. -100C and condensing saturation temperature of 300C. If clearance volume 15% of stroke volume. How will the volumetric efficiency vary for different refrigerant namely

(1) R-12 (2) R-22 (3)NH3? Solution: Volumetric efficiency can be given

suctionvol.

discharge

vn 1 C C

v

1) For R-12, from p-h chart 3

3

suction discharge

mv 0.07813 , v 0.025m / kg

kg

So vol.

0.07813n 1 0.05 0.05

0.025

= 0.894 = 89.4% For R-22, from p-h chart

3 3

suction dischargev 0.07m / kgv 0.0225m / kg

So vol.

0.07n 1 0.05 0.05

0.0225

= 1.05 – 0.0156 = 0.894 = 89.4%

(3) For NH3 , from p-h chart,

3

suction dischargev 0.4184m / kg,v

30.135m / kg

So vol.

0.4184n 1 0.05 0.05

0.135

= 0.895 = 89.5% Q.9 Catalogue data from R-12

compressor shows that compressor delivers 15 TR at saturation suction temperature of -50C and saturated condensing temperature of 400C and actual suction temperature is


150C. Liquid leaving the condenser is saturated. Calculate the mass flow rate of this compressor.

Solution:

from the p-h diagram

3

g1 gh 353KJ / kg,v 0.66m ,

3

1 1h 363KJ / kg,v 0.072m / kgP

The refrigeration effect

1 4m h h

15 210 m(363 239)

15 210m 25.40

124

m 25.40kg / min.

And volume flow rate V = 25.4 X 0.072 V = 1.828 m3/min.

Q.10 A geothermal wall at 1300C supplies

heat at a rate of a 100500 kJ/hr to an absorption refrigeration system. The environment is at 300C and the refrigerated space is maintained at -220C. Determine the maximum possible heat removed from refrigerated space.

Solution:

C E GT 303kT 251kT 403k

GQ 100500kJ / hr

For a reversible absorption cycle

G CE E

G C E G

T TQ TCOP X

Q T T T

251 403 303

303 251 403

E G

100500Q Q X1.19 X1.19

3600

EQ 32.22KW


Psychrometry is the branch of science which deals with study of the properties of the moist air i.e. mixtures of air and water vapour. Atmospheric air is a mixture of many gases plus water vapour and a number of pollutants. The amount of water vapour and pollutants vary from place to place. A mixture of various gases that constitute air and water vapour is known as moist air.

Constituent Molecular weight Mol fraction Oxygen 32 0.2095 Nitrogen 28.016 0.7809 Carbon dioxide 44.01 0.0003

7.1 PROPERTIES OF MOIST AIR

7.1.1 MOIST AIR

The moist air can be thought of as a mixture of dry air and moisture. Based on the above composition the molecular weight of dry air is found to be 28.966 and the gas constant R is 0.287035 KJ/Kg.K. If the partial pressure of dry air and water vapour is Pa and PV, then according to Dalton’s law of partial pressure

b a vP P P ------- (7.1)

Where bP is barometer reading of

atmospheric pressure. In general, atmospheric pressure is taken as 760 mm of Hg.

7.1.2 SATURATED AIR

At a given temperature and pressure, the dry air can only hold a certain maximum amount of moisture. When the moisture content in the moist air is maximum, the air is known as saturated air.

Fig. Water vapour in air mixture

7.1.3 HUMIDITY RATIO

The humidity ratio or specific humidity (W) is the mass of water per kg of dry air in a given mixture of dry air and water vapour. If mv is the mass of water vapour and ma is the mass of dry air. Then humidity ratio

v

a

W=m

m

We know that V V VP V m R Tforthewatervapour

And a a aVP V m R T for the dry air

Substituting the values

V

V V

a a V

P V

R T P RaW

P V P R

RaT

V V

a b V

P 8.314 / 28 P0.622

P 8.314 /18 P P

V

b V

PW 0.622 (7.2)

P P

If air is saturated at temperature T, then Specific humidity of saturated air

s

b s

PWs 0.622 (7.3)

P P

7.1.4 DEGREE OF SATURATION

The degree of saturation is the ratio of the humidity ratio (W) of the mixture to the humidity ratio of a saturated mixture (WS)at the same temperature and pressure. it is denoted by μ. Degree of Saturation

V

b V V b s

ss s b V

b s

0.622P

P P P (P P )Wμ

0.622PW P . P P

P P

V b s

s b V

P (P P )μ (7.4)

P . P P

7 PSYCHOMETRY


7.1.5 RELATIVE HUMIDITY Relative humidity is defined as the ratio of the mass of water vapour present in a given volume of air at temperature T to the mass of water vapour when the same volume of air is saturated at the same temperature T and pressure. Relative humidity is normally expressed as a percentage. When Φ is 100 percent, the air is saturated. Relative humidity

mass of water vapour in a given volume

of air at Temperature T

mass of water vapour when same volume of air is

saturated at Temperature T

V

S

m

m

Assuming the water vapour is an ideal gas,

V V VP V m R T

And for the saturated air condition

s s VP V m R T

So from the equation () and Equation (), we get

V V

S S

m P

m P

So V V

S S

m P

m P --------- (7.5)

for the dry air mV and PV are zero, so Φ = 0 for the dry air V Vm andP are equal to

s sm andP , so

Φ = 100% So relative humidity varies from 0% to 100%. 7.1.6 DRY BULB TEMPERATURE DBT is the temperature of the air as measured by a standard thermometer when it is unaffected by thermometer. 7.1.7 DEW-POINT TEMPERATURE (DPT) If moist air is cooled at constant pressure, then the temperature at which the

moisture presents in the air begins to condense is known as dew-point temperature (DPT) of air. 7.1.8 WET BULB TEMPERATURE (WBT) It is temperature of air measured by thermometer when wet cloth is covered over the thermometer. 7.1.9 DENSITY OF DRY AIR The density of air or air density is the mass of air per unit volume Air density decreases with increasing altitude. It also changes with variation in temperature or humidity. At sea level and at 15 °C, air has a density of approximately 1.225 kg/m3.density of dry air can be calculated using the ideal gas law, expressed as a function of temperature and pressure

aa

a D

T

6P

ρ 7R

.

Where ρ Air density

Pa = Absolute pressure TD = Dry Bulb temperature Ra = Specific gas constant for dry air

7.1.10 ENTHALPY OF THE AIR

The enthalpy of moist air is the sum of the enthalpy of the dry air and the enthalpy of the water vapour. Enthalpy values are always based on some reference value. The enthalpy of the mixture is given by

a V p a g W dph Wh C T W h 1.88 t th

------ (7.7) 7.2 PSYCHROMETRIC CHART A Psychrometric chart graphically represents the thermodynamic properties of moist air. Standard Psychrometric charts are bounded by the dry-bulb temperature line (abscissa) and the vapour pressure or humidity ratio (ordinate). The Left Hand Side of the Psychrometric chart is bounded by the saturation line. Figure shows the schematic of a Psychrometric chart.


http://en.wikipedia.org/wiki/Humidity

http://en.wikipedia.org/wiki/Sea_level

http://en.wikipedia.org/wiki/Kilogram_per_cubic_metre

http://en.wikipedia.org/wiki/Ideal_gas_law



http://en.wikipedia.org/wiki/Thermodynamic_temperature

http://en.wikipedia.org/wiki/Pressure

http://en.wikipedia.org/wiki/Specific_gas_constant

Psychrometric charts are readily available for standard barometric pressure of 101.325 KPa at sea level

and for normal temperatures (0-50oC).

The Relative humidity at the saturation curve is 100 %.

At the saturation curve, Dry bulb Temperature, Wet bulb Temperature and Dew point temperature are same.

fig. Psychrometric Chart The parameters of Psychrometric chart are relative humidity, dry bulb temperature, wet bulb temperature, dew point temperature, specific humidity, specific volume, specific enthalpy and vapour pressure. The Psychrometric chart allows all the parameters of some moist air to be determined from any three independent parameters, one of which must be the pressure. 7.3 SIGNIFICANCE OF DIFFERENT LINES ON PSYCHROMETRY CHART 7.3.1 DRY-BULB TEMPERATURE LINES

Dry bulb temperature is temperature of an air sample, as determined by an ordinary thermometer. It is typically plotted as the abscissa (horizontal axis) of the graph. The SI unit for temperature is degrees Celsius. These lines are drawn straight, parallel to each other. Each line represents a constant temperature.

7.3.2 WET-BULB TEMPERATURE (WBT) LINE Wet-bulb temperature is temperature of an air sample after the air has passed over a large surface of liquid water in an insulated channel. When the air sample is saturated with water, the WBT will read the same as the DBT. These lines are oblique lines that differ slightly from the enthalpy lines. They are identically straight but are not exactly parallel to each other. These intersect the saturation curve at DBT point.

fig. Wet Bulb Temperature line

7.3.3 DEW POINT TEMPERATURE (DPT) LINE Dew point temperature (DPT) is the temperature at which a moist air at the same pressure would reach water vapour saturation. At this point further removal of


http://en.wikipedia.org/wiki/Dry-bulb_temperature

http://en.wikipedia.org/wiki/X-axis

http://en.wikipedia.org/wiki/Degrees_Celsius



http://en.wikipedia.org/wiki/Wet-bulb_temperature

http://en.wikipedia.org/wiki/Wet-bulb_temperature

http://en.wikipedia.org/wiki/Dew_point

http://en.wikipedia.org/wiki/Dew_point

heat would result in water vapour condensing into liquid water fog. From the state point follow the horizontal line of constant humidity ratio to the intercept of 100% RH, also known as the saturation curve. The dew point temperature is equal to the fully saturated dry bulb or wet bulb temperatures.

fig. Dew Point temperature line 7.3.4 Relative humidity (RH) Line Relative humidity (RH) is the ratio of the mole fraction of water vapor to the mole fraction of saturated moist air at the same temperature and pressure. RH is dimensionless, and is usually expressed as a percentage. Lines of constant RH reflect the physics of air and water: they are determined via experimental measurement. Relative humidity: These hyperbolic lines are shown in intervals of 10%. The saturation curve is at 100% RH, while dry air is at 0% RH.

fig. relative humidity lines

7.3.5 HUMIDITY RATIO LINE Humidity ratio is the proportion of mass of water vapour per unit mass of dry air at the given conditions. It is also known as the moisture content or mixing ratio. It is

typically plotted as the ordinate (vertical axis) of the graph. For a given DBT there will be a particular humidity ratio for which the air sample is at 100% relative humidity. These are the horizontal lines on the chart.

Fig. Humidity ratio lines

7.3.6 SPECIFIC ENTHALPY LINE

These are oblique lines drawn diagonally downward from left to right across the chart that are parallel to each other. In the approximation of ideal gases, lines of constant enthalpy are parallel to lines of constant WBT.

Fig. Specific Enthalpy Lines

7.3.7 SPECIFIC VOLUME LINE Specific volume is the volume of the mixture containing one unit of mass of dry air. The SI units are cubic meters per kilogram of dry air. These lines are a family of equally spaced straight inclined lines that are nearly parallel.


http://en.wikipedia.org/wiki/Relative_humidity

http://en.wikipedia.org/wiki/Relative_humidity

http://en.wikipedia.org/wiki/Mixing_ratio

http://en.wikipedia.org/wiki/Mixing_ratio

http://en.wikipedia.org/wiki/Y-axis



http://en.wikipedia.org/wiki/Enthalpy

fig. Specific Volume lines

7.4 DIFFERENT PROCESS ON PSYCHROMETRIC CHART

7.4.1 SENSIBLE COOLING PROCESS

During this process, the moisture content of air remains constant but its temperature decreases as it flows over a cooling coil. For moisture content to remain constant the surface of the cooling coil should be dry and its surface temperature should be greater than the dew point temperature of air. If the cooling coil is 100% effective, then the exit temperature of air will be equal to the coil temperature. However, in practice, the exit air temperature will be higher than the cooling coil temperature. The heat transfer during the cooling process is given as

C a O A a pm O AQ m h h m C T T ----(7.8)

Fig. Sensible cooling Process 7.4.2 SENSIBLE HEATING PROCESS During this process, the moisture content of air remains constant and its temperature increases as it flows over a heating coil. The heat transfer rate during this process is given by

fig. Sensible Heating Process

h a b o a pm b OQ m h h m C T T (7.9)

Where Cpm is the humid specific heat. ma is the mass flow rate of dry air (kg/s). 7.4.2.1 BY PASS FACTOR In Fig., the temperature Td3 is the effective surface temperature of the heating coil, and is known as apparatus dew-point (ADP) temperature. Td1 and Td2 are the temperature of air at inlet and outlet of the coil. In an ideal situation, when all the air comes in perfect contact with the heating coil surface, then the exit temperature of air will be same as ADP of the coil.


But in actual case the exit temperature of air will always be lesser than the apparatus dew-point temperature. So By -pass factor (BPF) is defined as the ratio of heat loss to maximum possible heat transfer.

d3 d2

d3 d1

T THeat LossBPF

Maximum Possible heat transfer T T

--------- (7.10) In the same way, Bypass factor for cooling coil can be given as

d2 d3

d1 d3

T THeat LossBPF

Maximum Possible heat transfer T T

--------- (7.11) 7.4.2.2 EFFICIENCY OF COIL The efficiency of heating Coil is defined as the ratio of actual heat transfer to the cold air to maximum possible heat transfer to the cold air by heating coil. Efficiency

Actual Heat transfer

Maximum Possible heat transfer

d2 d1

d3 d1

T T

T T

-------- (7.12)

In the same way efficiency for cooling coil can be given as The efficiency of cooling Coil is defined as the ratio of actual heat transfer to the cold air to maximum possible heat transfer to the cold air by heating coil. Efficiency

Actual Heat transfer

Maximum Possible heat transfer

d1 d2

d1 d3

T T(7.13)

T T

By comparing the equation (7.11) and equation (7.13) of Bypass factor and efficiency of the heating and cooling coil, we get

1 BPF -------- (7.14)

7.4.3 COOLING AND DEHUMIDIFICATION PROCESS When moist air is cooled below its dew-point by bringing it in contact with a cold surface as shown in Fig., some of the water vapour in the air condenses and leaves the air stream as liquid, as a result both the temperature and humidity ratio of air decreases as shown. The heat and mass transfer rates can be expressed in terms of the initial and final conditions by applying the conservation of mass and conservation of energy equations.

Fig. Cooling

Fig. dehumidification By applying mass balance for the water ma.Wo = ma.WC + mw

By applying energy balance ma.ho = Qt +ma.hC + mw.hw

From the above two equations, the load on the cooling coil, Qt is given by:

Qt=ma(ho.-hC)+ma.(Wo-WC).hw ------- (7.15)

Where Wo and Wc are the humidity ration before and after the cooling process.mw is mass of water vapour separating from air. It can be observed that the cooling and de-humidification process involves both latent and sensible heat transfer processes, hence, the total, latent and sensible heat transfer rates (Qt, Ql and Qs) can be written as Qt = QL + QS

QL = ma.(Wo- WC).hw

QS = ma (ho.- hC) = ma.Cpm. (To.- TC)


7.4.3.1 SENSIBLE HEAT FACTOR

By separating the total heat transfer rate from the cooling coil into sensible and latent heat transfer rates, a useful parameter called Sensible Heat Factor (SHF) is defined. So SHF is defined as the ratio of sensible to total heat transfer rate.

SHFSensible Heat Transfer

Total Heat Transfer

=

Sensible Heat(SH)

Sensible Heat SH Latent heat(LH)

7.4.4 HEATING AND HUMIDIFICATION PROCESS During winter it is essential to heat and humidifies the room air for comfort. This is normally done by first sensible heating the air and then adding water vapour to the air stream through steam nozzles as shown in the fig. By this process, dry bulb temperature and humidity ratio increase. Appling Mass balance of water vapour for the control volume mw = ma (WD-Wo) Where ma is the mass flow rate of dry air.

Fig. Heating Process

Fig. Humidification Process

From energy balance: Qh = ma (hD-ho) - mw hw --------- (7.16)

Where Qh is the heat supplied through the heating coil. mw and hw is the mass and the enthalpy of steam. Since this process also involves simultaneous heat and mass transfer, we can define a sensible heat factor for the process in a way similar to that of a cooling and dehumidification process. 7.4.5 COOLING & HUMIDIFICATION PROCESS During this process, the air temperature drops and its humidity increases. This process is shown in Fig. This can be achieved by spraying cool water in the air stream. The temperature of water should be lower than the dry-bulb temperature of air but higher than its dew point temperature to avoid condensation (TDPT < Tw < TO). It can be seen that during this process there is sensible heat transfer from air to water and latent heat transfer from water to air. Hence, the total heat transfer depends upon the water temperature.

Fig. Cooling Process

Fig. Humidification process

If the temperature of the water sprayed is equal to the wet‐bulb temperature of air, then the net transfer rate will be


zero as the sensible heat transfer from air to water will be equal to latent heat transfer from water to air. The process is called as adiabatic dehumidification process.

If the water temperature is greater than WBT, then there will be a net heat transfer from water to air.

If the water temperature is less than WBT, then the net heat transfer will be from air to water.

7.4.6 CHEMICAL DE‐HUMIDIFICATION PROCESS

This process can be achieved by using a hygroscopic material, which absorbs or adsorbs the water vapour from the moisture.

Fig. Chemical De‐Humidification Process If this process is thermally isolated, then the enthalpy of air remains constant and the process is called as adiabatic dehumidification process. As a result the temperature of air increases as its moisture content decreases as shown in Fig. In general, the absorption of water by the hygroscopic material is an exothermic reaction, as a result heat is released during this process, which is transferred to air and the enthalpy of air increases. 7.4.7 MIXING OF AIR STREAMS Mixing of air streams at different states is commonly encountered in many processes, including in air conditioning. Depending

upon the state of the individual streams, the mixing process can take place with or without condensation of moisture. Fig shows an adiabatic mixing of two moist air streams. As shown in the figure, when two air streams at state points 1 and 2 mix, the resulting mixture condition 3 can be obtained from mass and energy balance.

fig. Mixing of two air streams From the mass balance of dry air and water vapour:

a1 a2 a3m m m

From energy balance:

a1 1 a2 2 a3 3 a1 a2 3m h m h m h m m h

a1 1 a2 2

3

a1 a2

m h m hh (7.17)

m m

From the above equation, it can be observed that the final enthalpy of mixture is weighted average of inlet enthalpies. Example:

Q.1 Atmospheric air at 760mm Hg pressure and dry bulb and wet bulb temperature150 C and 120 C enter the heating coil whose temperature is 410 C. The By pas factor of heating coil is 0.5. Determine the dry bulb temperature, wet bulb temperature, relative humidity of air leaving the heating coil and also the sensible heat supplied per kg of dry air. Solve the problem with help of psychrometric chart.

Solution: By pass factor of heating coil is given as


d3 d2

d3 d1

T TBPF

T T

d241 T0.5

41 15

0

d2 41 0.5 41 15 2T 8 C

Thus the actual condition at which air comes out of the heating coil is 280C.With the help of the psychrometric chart relative humidity at this point is 28.5% Wet bulb temperature at this point is 160C. From psychrometric chart, Enthalpy at point 1 h1 = 31.7 KJ/Kg Enthalpy at point 2 h2 = 44.9 KJ/Kg So the sensible heat supplied to the air Q =h2- h1

= 44.9-31.7 = 13.2 KJ/Kg of the dry air. Q.2 The humidity ratio of an

atmospheric air at 27.50C is 0.016 kg/ kg of dry air. determine partial

pressure of vapour, relative humidity and dew point temperature. Assume standard barometric pressure is 760 mm of Hg.

Solution: Specific humidity or humidity ratio is given by

V

b V

PW 0.622

P P

V

V

P0.016 0.622

760 P

0.016 760

Pv0.622 1.02575

= 19.05 mm of Hg. Relative humidity

V1

S1

HP

RP

Where Ps1 is saturated pressure corresponding to the saturation temperature equal to dry bulb temperature of 27.50 C. From the table it can be read as 27.535 mm of Hg.

So,

V1

S1

P 19.05RH 0.694

P 27.535

= 69.4 % Dew point temperature is the saturation temperature corresponding to the vapour pressure of the 19.05 mm of Hg as read from table. So, Tdp1 = 21.340 C

Q.3 Air enters a chamber at 5oC dry bulb temperature (DBT) And 2.5o C wet


bulb temperature at the rate of 3100m / min and the pressure of 1

bar. While passing through the chamber the air absorbs 50 KW heat and pick’s up 35 kg/hr of saturated steam at 110oC. Show the process on a psychrometric chart and find the dry and wet bulb temp of the leaving air at 110oC, enthalpy of saturated steam is 2691.3 kj/kg. Solution: For process 1-2 this is sensible heating process Heat absorbed = 50 kw

2 1m(h h )

i.e at state 1-2:

W1=0.0035(from psychometric chart)

Specific volume = 0.792m3/kg h1=14 kj/kg Volume flow rate =100 m3/min =1.67 m3/s

Mass flow rate1.67

0.792

= 2.104 kg/s Now heat transferred:-

2 1 2Q m( 2.104 m(h h ) h 14)

250 2.104 m(h 14)

h2= 37.76 kj/kg Now at state 2, from psychometric chart

2 2W 0.0035, h 37.76 kj / kg

After humidification by steam injection

3 2

msw w

mq

35

3600 2.10.0035 0.0081

Enthalpy of living air :

3 2 v

ms

mh h h

q

hv =2691.3 kj/kg at 110 0C

3

35

3600 2.10h 37.76 269

41.3

=50.19ks/kg At state (3):

t3 = 310C, tw2 =18.3 0C Q.4 In an air conditioned space, 50 kg

dry air/sec of fresh air at 45 0C DBT and 30% RH is introduced. The room air at 25 0C DBT and 50% RH is recalculated at 450 kg dry air/sec. The mixed air flows over a cooling coil which has apparatus dew point of 12 0C and by pass factor of 0.15. Determine the conditions at outlet of cooling coil, RSH, RLH, the cooling load of coil and the condensate rate .The saturation pressure of water of required temperature are

Temperature (0C) Pws (bar) 12 0.14016 25 0.03166 45 0.09584

Enthalpy at temperature, h=1.005 T +W (2500+1.885) kg/kg dry air.

Solution: Temperature of air after mixing

Tx = 0.9 t2 +0.1t1 = 0.9×25 +0.4× 45

Specific humidity: a) At outlet condition:

1W 0.622 Pv1

Po Pv1

Pv1 = 0.3×0.09584 = 0.02875 bar

1

0.02875

1.013W 0

2 0.02822

75.6

= 0.01816 kg vapour/kg dry air


b) In the Room

22

2

Pvw 0.622

Pv Pv

Pv2 = 0.5 ×0.03166 = 0.01583 bar

2

0.622 0.01583

1.0132 0.0 58W

1 3

W2 = 0.00987 kg vapour /kg dry air Specific humidity at A

A

0.622 0.014016W

0.0132 0.014016

= 0.008725 kg vapor /kg dry air Specific humidity at inlet to cooling coil

Wx = 0.9 W2+0.1 W1

= 0.9×0.009870 + 0.1×0.01816 =0.0107 kg vapour /kg dry air

Enthalpy at X hx=1.00×27+0.00107(25001.88+27)

= 54.43 kj/kg dry air Enthalpy at 2 H2=1.005×25 +0.00987(25001.88×25) = 50.3 KJ/kg dry air B.P.F. of cooling coil

ts ta

BPF(tx ta)

ts ta ts 12

0.15(tx ta) 27 12

0

sT 14.25 C

ws-wa ws-0.008725

=(wx-wa) 0.0107-0.00

0.15=8725

Ws = 0.009 kg vapour/kg dry air

Enthalpy at supply point

hs=1.005×14.25+0.009 (2500+1.88×14.25)

=37.06 KJ/kg dry air. Condition at outlet

Ts=14.25 0C hs =37.06 KJ/kg dry air

Ws=0.009 kg vapour/kg dry air Room sensible heat RSH = Ma Lp (T2-TS) =500×1.0216 (25-14.25 = 5491.1 KW Room latent heat RLH = Ma (w2-ws)+2500 =500 (0.00987-0.009) ×2500 = 1087.5 kw Load on cooling coil = ma (hx-hs) =500(54.43-37.06)

=8685 kw = 8685

3.5

= 2481.43 tones Condensate rate = (wx-ws) Ma

= (0.0107-0.009) ×500 =0.85 kg vapour /sec =3060 kg vapour/hr Q.5 Calculate all the psychometric

properties of air at 1 bar, 300C DBT and 250C WBT. The following properties of water may be assumed :-

Solution:

The following expression may be used, if necessary

v

P Pw DBT WBT 1.8

2824 1.325(1.8 DBT 3P

) Pv

2

DBT=250C, WBT=150C

P Pw DBT WBT 1.8

2854 1.325(P

1. Pv

8tDBT 32)

1 0.03166 25 15 1.8

2854 1.325(1.8 25 32

)0.017307

v17.43

2751.975 0.017307


Pv = 0.01097 bar Pyschrometric properties:- 1. specific humidity

W 0.622pv

p pv

0.01097

10.622

0.01097

W= 0.006899 kg vapor/kg of dry air 2. Relative humidity (ɸ)

V

Pv

P P100

0.01097100

0.03160

ɸ = 34.65% 3. Degree of saturation (µ)

S

V

P P)

Pµ

P(

(1 0.03166)0.3465

(1 0.00109)

0.968340.3465

0.9890

µ = 0.339 or 33.9% 4) Enthalpy of moist air (h)

h = Cpa + W [2500 + 1.88]

where Cpa =1.005 kj/kg.k

h 1.005 25 0.006899 2500 1.88 25

h = 25.125+0.006899(2547) h 42.696KJ / Kg of dry air .


Power plant engineering is the branch of thermal science which deals with study of chemical energy of fuel is converted into thermal energy and then utilises into the mechanical work done in the efficient manner. When the thermal energy is converted into the mechanical energy it is impossible to convert whole amount of heat into work. So efficiency play an important role in power plant to check the performance of the power plant. The main important power cycles are gas power cycle and steam power cycle. Gas power cycle works on Brayton cycle and steam power cycle works on Rankine cycle.

8.1 STIRLING CYCLE

Stirling cycle has two reversible isothermal heat additions & heat rejection process and two reversible isocoric processes. It differs from cannot cycle in that two isentropic processes of Carnot cycle are replaced by two constant volume process.

The efficiency of Stirling cycle is given as

3 min

1 max

T Tη 1 1

T T

If cycle is regenerative cycle then it has same efficiency as that of cannot cycle i.e.

area covered by 2 – 3 process on T-S diagram is equal to the area covered by 4 – 1 process.

8.2 ERICSON CYCLE

Ericson cycle has two reversible isothermal heat additions & heat rejection process and two reversible isobaric processes. It differs from cannot cycle in that two isentropic processes of Carnot cycle are replaced by two constant pressure process. P-V and T-s diagram of Ericson cycle is shown.

The efficiency of Stirling cycle is given as

2

1

T1-

Tη

Regenerative Ericson cycle has same efficiency as that at cannot cycle. And both Stirling and Ericson cycle has same efficiency it temperature limits are same. Gas turbine cycle with inter cooling, reheating & regeneration closely resemble the Ericson cycle. Both cycles are difficult in practice because heat transfer at constant temperature requires either infinite long surface area or infinite long time.

8.3 GAS POWER PLANT

8 POWER PLANT ENGINEERING


In gas power cycle, the gas is a working fluid. It does not go under the phase change during the cycle. Gas power cycle works on Brayton cycle. There are two type of cycle: Open Cycle In this cycle, the working fluid renew after completion of the cycle. Closed cycle In closed cycle, chemical composition of working fluid remains same as heat is supplied through the heat transfer and not by direct combustion. Advantages of closed cycle over open cycle 1) Use of high Pr. Throughout cycle

reduces size of plant 2) Close cycle has higher thermal

efficiency. 3) Elimination of possible of turbine blades

erosion 4) Closed cycle can use any working fluid

or gas of higher value of 𝛾 to increases the power output and efficiency.

8.3.1 BRAYTON CYCLE

Brayton cycle consists of four processes: Process 1–2: Isentropic Compression Process: In this process the heat transfer is zero. dQ = 0 Process 2–3 : Constant pressure Heat addition Process: Heat addition in this process is given by QS = m. CP T3 − T2 Process 3–4 : Isentropic Expansion Process: In this process the heat transfer is zero. dQ = 0 Process 4–1 : Constant pressure Heat rejection Process: The heat is reject in constant pressure process in heat exchanger. Heat rejection in this process is given by QR = m. CP T4 − T1

The P-V and T-S diagram of Brayton cycle has been shown.

The efficiency of Brayton cycle is given by

P 4 1R

S P 3 2

C T TQη 1 1

Q C T T

41

1

32

2

TT 1

T1

TT 1

T

------ (8.1)

process 1-2 and process 3-4 are isentropic process, so

γ 1

γ 1γ32 2 γ

1 1 4

TT Pr

T P T

Where r = compression ratio

3 4

2 1

T T

T T ------ (8.2)

substituting the values in equation 8.1, we get

1

2

Tη 1

T

1η 1

γ 1r

γ

Thus efficiency of Bray ton cycle depends upon compression ration and nature of gas.


8.3.2 ENERGY BALANCE EQUATION ON COMBUSTION CHAMBER

Heat released due to combustion of fuel is converted into enthalpy increases of gas in combustion chamber. Heat input = Enthalpy change of gas

f c.c g pg 3 2m XCVXη m c T T

f c.c a f pg 3 2m XCVXη m m c T T

8.3.3 ANALYSIS OF GAS TURBINE CYCLE The upper limit of temperature is fixed due to metallurgical condition of turbine blades. Though by increasing the Pressure

Ratio, net efficiency

γ 1

γp

11

R

can be

increased but net work will decrease. So there is a optimum limit of the compression ratio at which optimum efficiency and work output can be obtained.

Net work done in gas power plant is given by

net T Cw w w

P 3 4 2 1C T T T T

P 3 4 2 1C T T T T ----- (8.3)

form the process 1-2 and process 3-4

3 1 324

1 4 2

T T TTT

T T T

substituting the value in the equation 8.3

3 1net P 3 2 1

2

T Tw C T T T

T

for optimum condition differentiating the equation with respect to T2

net 3 1p 2

2

dw T T0 C 0 1 0

dt T

2 1 3T TT

1 34 1 3

1 3

T TT T .T

T T

4 1 3T T .T

Compression ratio for optimum work done

γ γ

γ 1 2 γ 132 2

1 1 1

TP Tr

P T T

γ

2 γ 13

1

Tr

T

net P 3 1 3 1 3 1w C T TT TT T

2

net max P 3 1w C T T

8.3.4 REGENERATION IN GAS TURBINE

Efficiency of gas can be increased by utilizing the energy of exhaust gases from the turbine exit in heating air leaving the comp. In heat exchanger, and this process is called as regeneration.

8.3.5 EFFECTIVENESS OF HEART TRANSFER It is ratio of actual temperature rise of air to maximum possible rise of the


temperature. It is also known as degree of regeneration.

a pa a 2

a f pa 4 2

m C T T

m m C T T

If the mass of air is higher as compare to mass of the fuel, so the effectiveness of heat exchanger is

a 2

m

4 2

T TActualtemp.Rise

max possibleRise T T

8.3.6 EFFECT OF REGENERATION 1) There is no change in turbine work. 2) There is no change in compressor work. 3) Net work done remains same 4) Heat Supplied is decreased so the

efficiency of the plant increase Efficiency of ideal regeneration cycle

b 1R

S 3 a

T TQη 1 1

Q T T

for the perfect regeneration,

4 2, a bT T T T

substituting the values in equation, we get

b 21 1

1 1

a 43 3

3 3

T TT 1 T 1

T T1 1

T TT 1 T 1

T T

γ 1

γ1

γ 1

3 γ

r 1Tη 1

T1 r

γ 1

1 γreg

3

Tη 1 r

T

So as compression ration increases the efficiency of simple Brayton cycle increases but the efficiency of regenerative Brayton cycle will decreases.

8.3.7 BRAY TON CYCLE WITH INTER COOLING The efficiency of Brayton cycle can be improved by use of stages compression with inter cooling. 8.3.7.1 EFFECT OF INTER COOLING 1) Decrease in compressor work 2) No change in turbine work 3) Network increases

Optimum Pressure Ratio for minimum compressor work

C C1 C2w w w

P 2 1 P 1 3C T T C T T

For perfect inter cooling 1 3T T

P 2 1 P 4 1C T T C T T

2 4P 1

1 1

T TC T 2

T T

γ-1 γ-1

γ γ2 i 4 2

1 1 1 i

T P T P= , =

T P T P

η η

i 2C P 1

1 i

P Pw C T 2

P P

Differentiating the equation with respect to Pi, we get

i 1 2p p p

substituting the value of pi in compressor work for perfect inter cooling

γ 1

γ2

C1 P 1

1

Pw C T 1

P


4C2 P 4 3 P 3

3

Tw C T T C T 1

T

γ 1

γ2

C2 P 1

1

Pw C T 1

P

8.3.8 BRAY TON CYCLE WITH REHEATING The efficiency of Brayton cycle can be increased by reheating the gas and passing it through the number of the turbines.

Effect of reheating 1) Turbine work increases because const

Pressure Line diverge 2) No change in compressor work 3) Network done increases Conditions for perfect reheating 1) 3 5T T

2) i 1 2P P P

3) T1 T2w w

4) T T1w 2w

8.3.9 ISENTROPIC EFFICIENCY OF COMPRESSOR AND TURBINE Due to irreversibility turbine produce less work as compare to ideal work and compressor requires more work as compare to ideal work required for compressor. efficiency of the compressor is given by

2 1 2 1C 1 1

2 1 2 1

h h T TIsentropic workη

actual work h h T T

efficiency of the turbine is given by

3 4 3 4T 1 1

3 4 3 4

h h T Tactual workη

Isentropic work h h T T

8.4 STEAM POWER PLANT In steam power plant, the working substance recalculates in two phases liquid and vapour. The components of simple vapour plant are shown in figure.

Simple vapour plant works on Rankine cycle. Heat is transferred to water in the boiler from external source. On P-h and T-S diagram, Rankine cycle can be shown.

Rankine cycle consists of four processes: Process1-2:IsentropicCompressionProcess: In this process woter is compressed isentropically in feed pump. Pump work can be given: Wp = h2 – h1

Process 2 – 3 : Constant pressure Heat addition Process in the boiler: The heat is added to water in the boiler. It is given as

3S 2hQ – h

Process 3 – 4 : Isentropic Expansion Process in the turbine: In this process the heat transfer is zero and work done by the turbine is given as

T 3 4 W h – h


Process 4 – 1 : Constant pressure Heat rejection Process in the condenser: The heat is rejected at constant pressure process in condenser. Heat rejection in this process is given by

4R 1hQ – h

The efficiency of Rankine cycle is given by

R

S

Q h4 – h1η 1 1

Q h3 – h2

8.4.1 REHEAT RANKINE CYCLE

In reheating cycle, the expansion of steam from the initial state at turbine is carried out in two or more no. of re heater. The network out increases and hence steam rate decreases.

netP m w

1) Work done by the turbine

T 3 4 5 6w h h h h

2) Work done by the turbine

p 2 1w h h vdp

3) Heat supplied in the boiler

S 3 2 5 4Q h h h h

T p

s

w wη

Q

The main advantage of reheat factor is to increase dryness fraction at the turbine exit.

8.4.2 REGENERATION IN RANKINE CYCLE

In order to increase the mean temperature of the heat addition so that heat can be transferred at higher temperature. So the efficiency of the cycle can be increased. In this cycle, it is possible by transferring the heat from vapour as it flows through the turbine to the liquid flowing around the turbine. With the help of open feed water heater (OFWH) heat is transferred between

the steam from turbine and water just before entering the boiler. Regenerative cycle can be shown on T-S diagram.

Turbine work in regeneration is given by

T 1 2 1 2 3w h h 1 x h h

Energy balance equation on OFWH

2 5 6xh 1 x h h

8.4.3EFFECT OF PARAMETERS ON PERFORMANCE OF STEAM POWER PLANT

1) Effect of Boiler pressure If the boiler pressure is increased, the effect on the performance can be checked with help of T-S diagram

1)

pw increases

2) Tw same

3) SQ Decreases

4) η increases 5) Dryness fraction at outlet of turbine

decreases

2) Effect of Superheating If steam is superheated in the boiler, the effect on the performance can be checked with help of T-S diagram


1)

pw same

2) Tw increases

3) netw increases

4) SQ increases

5) η iincreses

6) Dryness at the outlet of turbine increases

8.5 NOZZLE AND DIFFUSER 8.5.1 NOZZLE It is a device of varying cross section used to convert the Pressure Energy of working fluid into kinematic energy. It is mainly used to produce the jet of steam or gas of high velocity to produce throat for Jet propulsion to derive steam turbine or gas turbine.

8.5.2 DIFFUSER It is a device of varying cross section causing rise in Pressure Energy atthe cost of kinematic energy. These are used in centrifugal and axial compressor. The smallest section of nozzle is known

as throat.

8.5.3 SPEED OF SOUND It may be defined as velocity of Pressure Wave in fluid medium.

s

dpC

dp

8.5.4 VELOCITY OF SOUND IN IDEAL GAS For an ideal gas, flow is isentropic

γPv C

γ

PC

ρ

Taking log in both sides

g g gl p γl ρ l C

differentiating the equation, we get dp dρ

γ 0p ρ

S

dp γp

dρ ρ

2 γρRTC

ρ

So the velocity of sound in a medium is given by

C γRT

Where R is the specific Gas constant.

8.5.5 MACH NUMBER

Mach number is defined as the ratio of actual velocity to the velocity of the sound or sonic velocity.

VM

C

#M>1, then V>C flow is supersonic flow. # M = 1 , then V = C flow is Sonic flow. # M < 1 then V < C flow is Subsonic flow. # M>5 then V>5C flow is hypersonic flow. 8.5.6 STAGNATION POINT AND STAGNATION PROPERTIES Stagnation point is state in the fluid flow where velocity of flow is become Zero and kinetic energy converts into pressure energy. The Pressure, Density, temperature etc. of stagnation point are known as stagnation Properties.

for the ideal gas at stagnation point,

2

0

Vh h

2

2

P O P

VC T C T

2


2

O

P

VT T

2C

We know that P

γRC

γ 1

2

0

V γ 1T T

2 γR

2

0T V1 γ 1

T 2γRT

2

0

2

γ 1T V1 .

T 2 C

20γ 1T

1 MT 2

γ

γ 10 0P T

P T

8.5.7 FLOW THROUGH NOZZLE & DIFFUSER

Assumption for flow 1) flow is the isentropic flow. 2) Stagnation enthalpy is constant 0h c

2

0

Vh h

28.4

differentiating the equation dh VdV ------ (8.5) From combined first and second law of thermodynamic TdS dh VdP differentiating the equation, we get

dPdh VdP

ρ -------- (8.6)

or dP

Vdρ

dρρV

dV

dPdV

ρV -------- (8.7)

Continuity equation for the mass flow m ρAV

takinglogin both the sides

logm logρ logA logV

differentiating the equation

dρ dA dV0

ρ A V

2

dA dρ dP

A ρ PV

dA dV dρ V1 X

A V ρ dV

Substituting the value of dV from equation (8.7), we get

2

2 2

dA dP V1

A ρV C

2dA dVM 1

A V

1) When M < 1, flow is Subsonic flow

a) Converging passage dA

VeA

A V P M ρ

b) Diverging passage

dAVe

A

A V P M ρ

2) When M > 1 flow is supersonic flow

a) dA

VeA

A V P M ρ

b) Diverging Passage dA

VeA


A V P M ρ

Conclusion a) For subsonic flow and compressible

fluid convergent passage acts as a nozzle and divergent passage acts as a diffuser.

b) For supersonic flow and compressible fluid convergent passage acts as diffuser & divergent passage acts as a nozzle.

c) For an incompressible fluid converging passage always aits as nozzle and divergent as a diffuser.

ρ dA dV

0ρ A V

dA dV

A V

8.5.8 MASS FLOW RATE PER UNIT AREA mass flow rate per unit area is given by

2 n 1

n n1 P 1

2

mρ 2C T r r

A

For maximum flow rate

2

d m0

dR A

n

n 12

1

P 2r

P n 1

m

A

will be max when Area is minimum and

it occurs at throat. So the critical condition occurs at throat. So Critical pressure is given by

n

n 1c

1

P 2

P n 1

8.6 COMPRESSOR

A compressor is a device in which work is done on the gas to raise the pressure of the gas. The work done by the compressor is negative in the compression process because the work is done on the system to raise the pressure of the gas. 8.6.1 WORK DONE IN COMPRESSOR

Process of compression in compressor is shown in the fig. 1) Work done in adiabatic process

γ 1

γ2

1 1

1

PγP V 1

γ 1 PW

2) Work done in polytropic process

n 1

n2

1 1

1

PnW P V 1

n 1 P

8.6.2 ACTUAL DIAGRAM OF COMPRESSOR The actual work done will be more due to resistance offered by suction value and discharge value. The shaded portion represents the extra work to counter value resistance. The actual diagram with the clearance volume is shown in fig.

Fig.: Actual P-V diagram of compressor 8.6.3 CONDITION FOR MINIMUM WORK REQUIRED IN MULTISTAGE COMPRESSOR WITH PERFECT INTER COOLING For perfect inter cooling


1 3T T

2 1 3P P P

For multi stage compressor

32 4 n 1

1 2 3 n

PP P PC

P P P P

NN 1

1

PC

P

8.7 STEAM TURBINE Steam turbine is a prime mover which is used to convert the potential energy into the kinetic energy and then utilize it into the mechanical work. Main use of steam turbine is to produce the mechanical work in steam power plant. 8.7.1 TYPES OF TURBINE There are two types of turbines: 1) Impulse Turbine: In impulse turbine

pressure drops in nozzles but remains same in moving blade.

2) Reaction Turbine: In reaction turbine pressure drops in fixed blade and it also changes during the expansion of the steam in moving blades.

8.7.2 COMPOUNDING IN TURBINE In this type of compounding two shafts are used to drive separate generator at different RPM. There are two type of compound turbine: 1) Pressure compounding- Rateua turbine 2) Velocity compounding - Curtis turbine 8.7.3 VELOCITY TRIANGLE OF TURBINE

The velocity triangle is shown in the fig. From the outlet of nozzle, steam flow at absolute velocity V1. But due to blade velocity u, steam strikes at the blade with relative velocity Vr1.

Following are the parameters of the turbine. α=Angle at which steam enters or guide blade angle θ= Vane angle at inlet ϕ= Vane angle at outlet β= Angle at exit of moving blade

fig.: velocity triangle

1 2V &V = Absolute velocity at entrance and

exit

w1 w2V &V =Velocity of whirl at entry and at

exit or tan agent of absolute velocity r1 r2V &V Relative velocity at entry and

exital blade

f1 f 2V &V axial component of absolute

velocity u = blade velocity 8.7.4 PERFORMANCE PARAMETERS OF TURBINE

1) Velocity Ratio It is the ratio of blade velocity to the flow velocity at the inlet of the blade. It is given by ρ.

1

uρ

V

2)Blade or Diagram efficiency It is ratio of net work done by turbine to kinetic energy at the inlet of the blade.

w1 w2

D 2

1

2u V Vη

V

If a 2 1 2 1 1m T , then cos r R wθ V V V V


2 2 2

1 1

cos cos

cos cosα 2

w r rV V u V u

V u u V u

1 2 1

2 2

1 1

2u 2u 2 coscosα u

V

w w

D

V V Vη

V

2

D 2

1 1

4ucos uη 4

V V

24ρcosα 4ρ

Dη 4ρ cosα ρ --- (8.8)

For maximum diagram efficiency, differentiating the equation with respect to ρ.

Ddη0 4cosα 8ρ

dp

cosαρ

2

Substituting the value of ρ in equation 8.8, get

2

Dmax

cos asη 4 as cos

2 2

2

Dmaxη cos

For n number of stages (Curtis stage)

opt

cosαρ

2η

2

Dmaxη cos α

3)Blade or diagram efficiency of reaction turbine

2

D 2

2 2ρcosα ρη

1 2ρcosα ρ

For maximum diagram efficiency, differentiating the equation with respect to

, we get

Ddη0

dρ

ρ cosα

Substituting the value of in equation of diagram efficiency of reaction turbine, we get

2

Dmax 2

2 2cos αη

1 cos α

4) Stage efficiency It is the ratio of the net work done to change of enthalpy at the inlet of the turbine.

w1 w2

stage

1 2

u V Vη

h h

Examples Q.1 Determine the minimum no. of stage

required in an air comp. which takes air at 1 bar 270C and delivers 180 bar. The max discharge temperature is 1500C and considers the index of polytrophic as 1.25 and perfect inter cooling b/w stages.

Solution:

NN 1

1

PC

P

1.25

1.25Nn 1 2

1 1

P T5.57

P T

N1805.57

1

N= 3 Q.2 A gas turbine plant operates on

Brayton cycle, between Tmin =300K and Tmax=1073K. Find the maximum work done per kg of air and cycle efficiency. Compare the efficiency with the efficiency of Carnot cycle with same temperature limit.

Solution:

netmax

2

Max MinTC TW p

2

107 1. 0005 3 3 0

= 239.28 KJ/Kg Efficiency of the Brayton cycle

γ 1

γ

1η 1

r

Min

Max

T 3001 1 47%

T 1073

Efficiency of the Carnot cycle

minC

max

T 300η 1 1

T 107 72.1%

3

C

η 0.470.65

η 0.721


Q.3 A gas turbine plant draws in air at 1.013 bar, 10°C and has a pressure ratio of 5.5. the maximum temperature in the cycle is limited to 750°C. Compression is conducted in rotary compressor having an isentropic efficiency of 82%, and expansion takes place in a turbine with an isentropic efficiency of 85%. A heat exchanger with an efficiency of 70% is fitted between the compressor outlet and combustion chamber. For an air flow of 40 kg/s, find:

a) The overall cycle efficiency, b) The turbine output, and Solution: P1=101.3 KPa, T1=283 K, r=5.5,

T4 = 750℃ = 1023 K

γ 1

γ 1γ2 2 γ

1 1

T Pr

T P

1.4 1

1.45.5

T2 = 460.6 K

Isentropic efficiency of compressor

2 1 2 1C 1 1

2 1 2 1

h h T TIsentropic workη

actual work h h T T

2 1

1 1

2 1 2

T T 460.6 2830.82

T T T 283

T2’ = 499.6K

γ 1 1.4 1

4 γ 1.4

5

r 5.5

T

T

T5 = 628.6 K

Efficiency of turbine

4 5T 1

4 5

T Tactualworkη

Isentropicwork T T

1

5

1023 638.60.85

1023

T

T5’ = 687.7 K

Maximum possible heat from heat

exchanger = Cp(T5 –T2)

Actual Heat transfer = 0.7Cp(T5 –T2)

= 132 KJ/kg of air

Cp(T3-T2) = (1+ m)×132

CpT3 = 634.43 +132.33 m

m = 8.68 × 10–3 kJ/kg of air

Heat addition (Q1)=Cp(T4-T3)=CpT4-

CpT3

= 393.7 –132.33m

= 392.6 kJ/kg of air

WT = (1 + m) (h4 –h5)

= (1 + m) Cp (T4 – T5’)

=1.00868 × 1.005 × (1023 – 687.7)

kJ/kg of air

= 340 kJ/kg

Wc=(h2–h1)=Cp(T2‘–T1)=1.005×

(499.6 – 283)

= 217.7 kJ/kg of air

Wnet = WT - Wc = 122.32 KJ/kg

a) Networkdone 122.32

Heatsupplied 392.6 31.16%

b) Turbine output = (WT) = 122.32

kJ/kg of air= 4893 kW

Q.4 A simple steam power cycle uses

solar energy for the heat input. Water in the cycle enters the pump as a saturated liquid at 40°C, and is pumped to 2 bar. It then evaporates in the boiler at this pressure, and enters the turbine as saturated vapour. At the turbine exhaust the conditions are 40°C and 10% moisture. The flow rate is 150 kg/h. determine: a) the turbine isentropic efficiency, b) The net work output c) The cycle efficiency

Solution From Steam Table

T1 = 120.23°C = 393.23 K, h3 =

2706.7 kJ/kg, s3 = 7.1271 kJ/kg – K

At 40°C saturated pressure 7.384

kPa


hf=167.57KJ/Kg, hfg=2406.7 KJ/Kg

sf = 0.5725, sg = 8.2570

h4’ = hf + 0.9 × 2406.7

= 2333.6 kJ/kg

For h4s if there is dryness fraction x

7.1271=0.5725+x×(8.2570– 0.5725)

x = 0.853

h4 = 167.57 + 0.853 × 2406.7

= 2220.4 kJ/kg

Isentropic efficiency,

isentropic

h3 – h4 2706.7 – 2333.6

h3 – h4' 2706.7 – 22 0

2 .7

= 76.72%

Net Turbine work output WT=h3 – h4

= 373.1 kJ/ kg

Pump work, WP = v (P1 – P2)

= 1.008 × 10–3 (200 – 7.384) kJ/kg

= 0.1942 kJ/kg

Power = m × (W T - Wp) = 15.55 kW

h3=2706.7kJ/kg, h2=167.76 kJ/kg

Q1=(h3–h2)=(2706.7 – 167.76) kJ/kg

= 2539 kJ/kg

cycle T P

1

W W 373.1 0.1942

Q 2539 14.69 %

Q.5 A regeneration Rankin cycle as steam entering turbine at 200 bar and 250℃ superheated state and leaving at 0.05 bar. Considering feed water to be open type. Determine the efficiency of plant when feed water is operating at 8 bar.

Solution: At 200 bar and 250℃

h=3675.2KJ/Kg, s=6.6502KJ/Kg.K,

𝑣𝑓 = 0.001005 m3/Kg

At 0.005 bar hf = 137.82 KJ/Kg,

hfg = 2423.7KJ/Kg,

KJsf 0.4764 .K sfg 7.9187KJ / Kg.K

Kg

vf = 0.001005 m3/ Kg

At 8 bar

hf = 721.11KJ/Kg hfg = 2048KJ/Kg,

f fg

KJs 2.0462 .Ks 4.6166

Kg

KJ/Kg.K

vf = 0.001115 m3/ Kg

Regeneration cycle can be shown on

T-s diagram

Process 1-2 is isentropic process

s1 = s2 , h1 = 3675.2

6.6502 = 2.0462 + 𝑥2 4.6166

𝑥2 = 0.997

Enthalpy of point 2

2 f 2 2 fg2h h x h 721.11 0.997 2048

2h 2763.54kJ / kg

Process 1-3 is isentropic process

𝑠1 = 𝑠3

6.6502 = 0.4764 + 𝑥3 𝑋 7.9187

𝑥3 = 7796

Enthalpy of Point 3

3 3 3 3 f fgh h x h

137.82 0.7796 2423.7 X

2027.45 / kJ kg

Work done by pump 1

2

p1 5 4w =h -h =vdp=0.001005 8-0.05 ×10

=0.798

5h =0.798+137.82=138.618kJ/kg

Work done by pump 1

2

p2 7 6w =h -h =0.001115 200-8 X10

= 22.08 𝑘𝐽/𝑘𝑔

7 6h =h +22.08=723.19kJ/kg


Energy Balance equation at open

feed water heater

2 5 6x h 1 x h h

2763.54 1 138.618 721.11 x x

x = 0.2919

net 1 2 2 3

p1 p2

w h h 1 x h h

w w

3675.2 2763.54 1 0.2919

2763.54 2027.45 (798 22.08)

1462.4KJ / kg

s 1 7Q =h -h =3675.2-723.19

= 2932.78 kJ/kg Efficiency of the plant

net

s

wη=

Q

1462.449.86%

2932.78

Q.6 A stream of air flow in a duct of

100mm diameter at the rate of 1 kg/sec. The stagnation temperature is 370C. At one section of the duct the static pressure is 40 KPa. Calculate the Mach number, velocity and stagnation pressure at that section.

Solution: T0 = 37+ 273 =310 K, P= 40 KPa, and γ = 1.4 The mass flow rate per unit area is

m P

ρ V M γRTA RT

2

0

0 0

Tγ PM γ PM (γ-1)M= × = × 1+

R T R 2T T

2

2

1 1.4 40M (1.4-1)M= × 1+

π0.1 0.287 2310

4

3 2

3

1.4127.39 40 10 1 0.2

0.287 10 310

M M

0.803 = 2M 1 0.2M

Squaring both the sides

0.645 = 2 2M (1 0.2M )

0.2 M4 + M2 − 0.645 = 0 Solving the equation, we get M = 0.76

2 2

0T (γ 1)M (1.4 1) 0.761 1

T 2 2

T = 277.78K

C= γRT

= 31.4 0.287 277.78 10

= 334.08 m/sec V = c × M = 334.08×0.76 = 253.9 m/s

γ 1 1.4 1

γ 1.40 0P T 310

P T 277.78

P0 = 40 × 1.468 = 58.72 KPa Q.7 A simple open cycle gas turbine has

a compressor turbine and a free power turbine. It develops electrical power output of 250 MW .the cycle takes in air at 1 bar and 288 K. The total compressor pressure ratio is 14. The turbine inlet of compressor is 1500 K. The isentropic efficiency of compressor and turbine is 0.86 and 0.89 respectively. The mechanical efficiency of each shaft is 0.98. Combustion effect is 0.98 while combustion pressure loss is 3% of compressor delivery pressure. Alternator efficiency is 0.98. Take calorific value of fuel is 42000 KJ/kg. Cpa = 1.005 KJ/Kg K and Cpg = 1.15 KJ/Kg. K. Calculate i) Air –fuel ratio ii) Specific work output

Solution: Electrical power =250mw Power output from power turbine

250

alternator efficiency mechanical efficiency

250

0.98 0.98

= 260.3082 MW

= 260308.2 KW

0.4

1.288 14 612. K

415

’

2 1 2 2T T 14 T T1 21 1)

T T

ηc


612.15 288

0.86288 664.92K

Compressor is run by compressor turbine MaCpa(T2 - T1) = ma (1 + f) (T4 –T5 ) cpg × 0.98

1.005 × (664.92-288) = (1+t) (tut5) × 1.15 × 0.98

4 5

250 329.4

0.98 98 0.1

98f T – T

xo.

(1+F)(1500-T5) = 336.12 --------- (1) Total efficiency of turbine = 0.89

1.331.33

64

1.33 1 1.33 1

4 6

TT

P P

0.331 1.33

1 66 4

4

PT T

P 791k

4 6 6

1

6 6

0.89 T T 1500 T

(T T ) 1500 791

=

T6 = 1500 – 0.89 (1500 – 791) T6 = 869K

Power output from power turbine, Ma(1+F) Cpg (T5 − T6) = 260308.2

a 5M T 81 F 691.15 260308.2

Ma(1+F) (T5 − 869) = 226355 Heat balance equation in turbine,

a 4 a 2M T M1 F Cpg Cpa T

aM CV combustion effie Fncy 1

1.15 1 500 –1.005 664.92 41160 f

1725 – 668.25 = (41160 – 1725) F F = 0.0268

i) Air fuel ratio = 1

F = 37.32

ii) Specific work output Substituting the value of F in equation 1

(1+0.0268) (1500 - T5) = 336.12 T5 = 1500 – 327.35 T5 = 1172.65 K Ma (1+0.0268) (T5 − 869) = 226355 Ma = 725.987 Ma = 726 kg/sec Special output =(1+F) Cpg(T5 − T6)

= 1.0268 1.15 (1172.65 – 869)

= 358.56 KW/Kg Q.8 In a multi-stage parson’s reaction

turbine at one of the stages the rotor diameter is 125 cm and speed ratio 0.72. The speed of the rotor is 3000rpm. Determine 1) The blade inlet angel if the blade

outlet angle is 22°

2) Diagram efficiency Solution:

Parson’s reaction turbine D1=1.25m, N=300rpm ρ = u/v1=0.72m, β1=?, β2=22 For parson’s reaction turbine α1 =β2 and α2= β1

π D N π 1.25 3000

60 6u

0

= 196.34 m/s V1 = u/V1 = u/0.72 = 272.69 m/s Vr2 = V1 = 272.69 m/s

Inlet velocity triangle and Outlet velocity triangle

Inlet velocity triangle outlet velocity triangle

Vw1 = 272.69 cos20 = 252.83 m/s Vf1= 272.69 sin20 = 102.15 m/s


1

Vf1 102.15

vw1 vtan

1 25

2.

83 196.34

So, β1=61.05 and α2=β1-61.05 diagram efficiency

2

d2 2

power

1m[V1 V2 Vf1 ]

2

power = mu (vw1+vw2) u = 196.34 m/s vw1 = 252.83 m/s

From outlet velocity triangle vr2 cos β2 = u+Vw2 272.69 cos 22=196.34+Vw2 Vw2=56.49 m/s

Power = 1 × 196.34 × (25 + 56.49) = 60.73KW

3

d2 2 2

60 10

1m[272.69 272.69 102.15 ]

2

0.90or 90 %

Q.9 A de level turbine has a mean blade speed of 180 mps. The nozzles are inclined at 170 to the tangent. The steam flow velocity through the nozzle is 550 m/sec. For a mass flow of 3300 kg/hour and for axial exit condition: Find 1) The inlet and outlet angles of the

blade system. 2) The power output. 3) Diagram efficiency.

Solution: u= 180 m/s Nozzle angle αͦ = 170, Mass flow rate

= 3300 kg/hr = 0.9167 kg/sec V1 =550 m/s Inlet velocity triangle and outlet

velocity triangle

Inlet velocity triangle Outlet velocity

triangle from velocity triangle Vw1= v1 cosα= 550 cos17 =525.96

m/s Vf1= v1 sinα = ssosinl7=160.80 m/s

1

vf1

vw vt n

1a

1

1

160.80

525.96tan

180

β1 = 24.92

2T n

va

vf 2

2

2

vf1t

160.80

u 18n

0a

β2 = 41.27°

Power output = m (Vw1+Vw2) u vw1 =. 525.96 m/s vw2=0 (due to axial outlet condition) u = 180 m/s

P 525.330

960

36000 180

P = 86.78 KW iii) Diagram efficiency

2

1

2(vw1 vw21

)0

u

V0

2 (525.96 0) 180

(550

21 0

0)

= 62.59 % Q.10 A single acting two-stage air

compressor deals with 4 m3/min of air at 1.013 bar and 15℃ with a speed of 250 RPM. The delivery pressure is 80 bar. Assuming complete inter -cooling, find the power required by compressor and bore and stroke of the compressor. Assume apiston speed of 3 m/s, mechanical efficiency 75% and volumetric efficiency of 80% per


stage. Assume the polytropic index of compression in both stage to be n = 1.25 and neglect clearance.

Solution: Intermediate pressure for perfect intercooling

P2 = P1P3 = 1.013 × 90 = 9 bar

Minimum power required by compressor

W =

n 1

n2

1 1

1 mech

P2n 1P V 1

n 1 P η

1.25 1

1.252 1.25 1.013 100 4 91

1.25 1 0.75 60 1.013

1013 40.548 49.34

45

KW

L is the length of stroke of the piston

2L 3 m / sN

60

90 100

25L 3

06 cm

Effective swept volume Vs = 4/ 250 = 0.016 m3

2

LP Vol

π×D ×L×η =0.016

4

LP

0.016 4D

3.14 0.36 0.8

= 0.266 m = 26.6 cm Ideal gas equation

3 31 1

1 3

P VP V

T T

for perfect intercooling, T1 = T3 So.

31

3 1

PV

V P

2

LP

2

HP

πD L

94π 1.013

D L4

HP

1.013D 0.266

9

= 0.89 m = 89 cm


9.1 HEAT ENGINE

Heat engine is a device which transforms the chemical energy into heat energy and the utilized it into mechanical work in an efficient manner. Thus the thermal energy is transforms into mechanical energy into heat engine. There are two types of heat engine:

The internal combustion engine is a heat engine that converts heat energy (Chemical energy of a fuel) into mechanical energy inside the cylinder of the engine (usually made available on a rotating output shaft). The External combustion engine is a heat engine that converts heat energy (Chemical energy of a fuel) into mechanical energy outside the cylinder of the engine. The internal combustion engine can be categorized into two types:

1) Spark Ignition (SI) Engine

An SI engine starts the combustion process in each cycle by use of a spark plug. The spark plug gives a high voltage electrical discharge between two electrodes, which ignites the air fuel mixture in the combustion chamber surrounding the plug. In early engine development, before the inventor of electric spark plug, many forms of torch holes were used to initiate combustion from an external flame.

2) Compression Ignition (CI) EngineThe combustion process in a CI engine starts when the air-fuel mixture self-ignites

due to high temperature in the combustion chamber caused by high compression.

9.2 APPLICATIONS OF IC ENGINES

Mainly used as ‘prime movers’, e.g. for be the propulsion of a vehicle car, bus, truck, locomotive, marine vessel, or airplane. Other applications include stationary saws, lawn mowers, bull-dozers, cranes, electric generators, etc.

9.3 CLASSIFICATIONS OF IC ENGINES

IC engines can be classified according to: 1. Number of cylinders – 1, 2, 3, 4, 5, 6 to

16 cylinder engines.2. Arrangement of cylinders – Inline, V-

type, Flat type, etc.3. Type of cooling – Air-cooled, Water-

cooled, etc.4. Number of strokes per cycle – 2-stroke,

4-stroke engines.5. Method of ignition – Spark Ignition (SI),

Compression Ignition (CI).6. Primary mechanical motion –

Reciprocating, rotary.

9.4 BASIC COMPONENTS OF I.C. ENGINE

Fig: Internal Combustion Engine

9 INTERNAL COMBUSTION ENGINE


Engine components: The following is the list of major components found in most reciprocating internal combustion engines

Block: Body of engine containing the cylinders made of cast iron or aluminium. In many older engines the valves and the valve ports were contained in the block. On air cooled engines the exterior surface of the block has cooling fins.

Camshaft: Rotating shaft used to push open valves at the proper time in the engine cycle either directly or through mechanical or hydraulic linkage. Most modern automobile engines have one or more camshafts mounted in the engine head.

Carburettor: Venturi flow device that meters the proper amount of fuel into the air flow by means of pressure differential.

Catalytic converter: Chamber mounted in exhaust flow containing catalytical material that promotes reduction of emission by chemical reaction.

Choke: Butterfly valve at carburettor intake, used to create rich fuel-air mixture in intake system for cold weather starting.

Combustion chamber: The end of the cylinder between the head and the piston face where the combustion occurs.

Piston: The cylindrically shaped mass that reciprocates back and forth in the cylinder transmitting the pressure forces in the combustion chamber rotating the crank shaft. The top of the piston is called crown and the sides are called skirt. Pistons are made of cast iron Aluminium or steel.

Piston rings: Metal rings that fit into circumferential groups around the piston and form a sliding surface against the cylinder walls. The purpose of the rings is to form a seal between the piston and cylinder walls.

Flywheel: Rotating mass with large moment of inertia connected to the

crankshaft of the engine. The purpose of the flywheel is to store energy and furnish a large angular momentum that that keeps the engine rotating between power strokes and smoothes out engine operation.

Fuel pump: Electrically or mechanically driven pump to supply fuel from the fuel tank to the engine. 9.5 TERM SUSEDIN INTERNAL COMBUSTION ENGINE

i) Top-Dead-Centre (TDC): Position of the piston when it stops at the furthest point away from the crankshaft.

ii) Bottom-Dead-Centre (BDC): Position of the piston when it stops at the point closest to the crankshaft.

iii) Bore: Diameter of the cylinder or diameter of the piston face, which is the same minus a very small clearance.

iv) Stroke: Movement distance of the piston from one extreme position to the other: TDC to BDC or BDC to TDC. It is denoted by L.

v) Clearance volume: Minimum Volume in the combustion chamber with piston at TDC. It is given by Vc.

vi) Displacement volume: Volume

displaced by the piston as it travels through one stroke. It is also known as swept volume (Vs).

vii) Air Fuel Ratio: It is the ratio of mass

air to mass of fuel input into engine. viii)Stroke to Bore Ratio: It is the ratio of

the stroke length to the diameter of the cylinder.

L

Stroke to bore ratio=d

If L

1d It is known as Square Engine

L

1d It is known as Under Square

engine


L

1d It is known as Over Square

engine ix) Compression ratio: It is the ratio of the

total volume to clearance volume.

C S STP

C C C

V V VVr 1

V V V

9.6 DIFFERENCE BETWEEN FOUR STROKE AND TWO STROKE ENGINE

9.7 PERFORMANCE PARAMETERS

1) Indicated thermal efficiency: it is the ratio of the energy in the indicated power to input fuel energy in the appropriate unit.

ith

I.P.(KJ / Sec)η

energy of fuel / sec

2) Brake thermal efficiency: it is the ratio of the energy in the brake power to input fuel energy in the appropriate unit.

ith

B.P.(KJ / Sec)η

energy of fuel / sec

3) Mechanical efficiency: it is the ratio of the energy in the brake power to indicated power in the appropriate unit.

m

B.P..(KJ / Sec)η

I.P..(KJ / Sec)

4) Friction Power: friction power is the

difference between indicated thermal power and mechanical power or brake power.

F.P. =I.P.-B.P.

5) Relative efficiency: it is the ratio of the actual thermal efficiency to air standard

efficiency.

r

actually thermal eff.η

Air standard efficiency

6) Mean effective pressure: M.E.P. is

average pr. inside the cylinder of an internal combustion engine on power output

imP X A X L X NXKI.P. watt

60

Where N

N for4 stroke engine2

and

N N for 2 – stroke engine the mean effective pressure can be given as

im

area of indicator diagramP

Length of indicator diagram

9.8AIR STANDARD CYCLE AND EFFICIENCY

The analysis of actual cycle is very complicated. So the analysis of cycle is done on the basis of air standard cycle. The air standard cycle is based on such assumptions: I. The working medium is perfect gas. II. There is no change of the mass of the

working fluid. III. All the processes in the cycle are

reversible processes.

9.8.1 OTTO CYCLE

Nicolaus Otto proposed a constant volume heat addition and rejection process in place of isothermal process of Carnot cycle. Otto cycle is used now a day in spark ignition engine. 4-stroke petrol engine operates on air standard Otto cycle. It completes the Otto cycle in 4 strokes (4 TDC to BDC movements of the piston). In 4 stroke crank shaft rotates 7200. The four stroke of four stroke engine are: 1) Suction Stroke, 2) Compression Stroke, 3) Power Stroke, 4) Exhaust Stroke. Otto cycle shown below consists of four processes:


Process 1–2: Isentropic Compression Process: In this process the heat transfer is zero. dQ = 0 Process 2 – 3 : Constant Volume Heat addition Process: The fuel and air mixture is burnt at constant volume process. Heat addition in this process is given by

S v 3 2Q m.C T T

Process 3–4 : Isentropic Expansion Process: In this process the heat transfer is zero. dQ = 0 Process 4–1 : Constant Volume Heat rejection Process: The heat is reject in constant volume process. Heat rejection in this process is given by

R v 4 1Q m.C T T

P-V and T-S diagram for Otto engine is shown in the fig.

fig: Otto cycle: P-V and T- S diagram Efficiency of the cycle is given by

R

S

Qη 1

Q

V 4 1

V 3 2

m.C T T1

m.C T T

4 1

3 2

T T1

T T

41

1

32

2

1

1

1

TTT

TTT

----- (9.1)

process 1-2 is isentropic process so for the isentropic process,

γ 1

γ 12 1

1 2

T Vr

T V

Process 3-4 is the isentropic process so for the isentropic process,

γ 1

γ 13 4

4 3

T Vr

T V

1 4

2 3

V Vr

V V

32

1 4

TT

T T

Substituting the value in equation 9.1

4

11 1

2 23

2

T1

TT Tη 1 1

T TT1

T

1

11

r ------------ (9.2)

The thermal efficiency is the function of the compression ratio and the ratio of specific heat γ. if the is considered as constant the

efficiency depends upon the compression ratio. The effect on efficiency of the Otto cycle can be understood with the help of the efficiency vs. compression ratio.

fig: efficiency vs. compression curve 9.8.2 DIESEL CYCLE

In actual spark ignition engine, upper limit of compression ratio is limited due to self-ignition temperature of the petrol. This can be overcome by compressing the air separately. In diesel cycle fuel is burnt at constant pressure. Other three processes are the same processes as in Otto cycle.


Diesel cycle shown below consists of four processes: Process 1–2: Isentropic Compression Process: In this process the heat transfer is zero. dQ = 0 Process 2 – 3 : Constant pressure Heat addition Process: The fuel is burnt at constant pressure process. Heat addition in this process is given by

S P 3 2Q m.C T T

Process 3–4 : Isentropic Expansion Process: In this process the heat transfer is zero. dQ=0 Process 4–1 : Constant Volume Heat rejection Process: The heat is rejected in constant volume process. Heat rejection in this process is given by

R v 4 1Q m.C T T

P-V and T-S diagram for diesel engine is shown in the fig.

Efficiency of the cycle is given by

R

S

Qη 1

Q

V 4 1

D

P 3 2

4 1

3 2

m.C T Tη 1

m.C T T

1 T T1 (9.3)

γ T T

Compression Ratio 1

2

Vr

V

Cut off ratio 3C

2

Vρ

V

Expansion ratio 4e

3

Vr

V

4 2

3 2

V V

V V

1 2

2 3 C

V V r

V V ρ

C er p r

Substituting the values of the temperature in term of the compression ratio and cut off ratio, we get efficiency of diesel engine

γ

cD γ 1

c

ρ 11η 1

r γ ρ 1

----------- (9.4)

The bracket term is always greater than one, so efficiency of diesel cycle is always less than efficiency Of Otto cycle for given compression ratio. The normal range of compression ratio in petrol engine is from 6–10 and in diesel 15–20. So efficiency of diesel engine is more than efficiency of gasoline engine.

9.8.3 DUAL CYCLE

Dual cycle is also called as mixed cycle or limited pressure cycle. It is the compromise between the Otto cycle and Diesel cycle. Because for chemical reaction some time is required to so combustion does not occurs at constant volume and due to rapid uncontrolled combustion in diesel engines, combustion does not occur at constant pressure. Dual cycle shown below consists of five processes: Process1-2: Isentropic Compression Process: In this process the heat transfer is zero. dQ = 0 Process 2–3 : Constant volume Heat addition Process: The fuel is burnt at constant volume process. Heat addition in this process is given by

S1 V 3 2Q m.C T T

Process 3–4: Constant pressure Heat addition Process: The fuel is burnt at constant pressure process. Heat addition in this process is given by


S2 P 4 3Q m.C T T

Process 4–5 : Isentropic Expansion Process: In this process the heat transfer is zero. dQ = 0 Process 5–1: Constant Volume Heat rejection Process: The heat is rejected in constant volume process. Heat rejection in this process is given by

R v 5 1Q m.C T T

P-V and T-S diagram of Dual cycle is shown in the fig.

Efficiency of the Dual cycle is given by

R

S

Qη 1

Q

v 5 1

p 4 3 v 3 2

m.c T T1

m.c T T m.c T T

5 1

4 3 3 2

T T1

γ T T T T

Compression Ratio 1

2

Vr

V

Pressure ratio 3

2

Prp

P

Cut off ratio 4C

3

Vρ

V

Expansion ratio 5 5 3e

4 4 3

V V Vr

V V V

31

2 4 C

VV r

V V ρ

er c r

Substituting the values of temperatures in the term of compression ratio, pressure ratio and cut off ratio, we get Efficiency of Dual cycle

γ

p c

DUAL γ 1

p p c

r ρ 11η 1

r (r 1) γr ρ 1

--- (9.5)

9.9 COMPARISON OF OTTO, DIESEL, DUAL CYCLE 1) Same comp. ratio and heat addition

Otto cycle 1-2-3-4, Diesel cycle 1-2-3’-4’ and Dual cycle 1-2-2’-3”-4” are shown on P-V and T-S diagram. From the T-S diagram it is clear that heat input i.e. area of T-S curve along S- axis is same.

But the heat rejected in Otto cycle is minimum and in Diesel cycle is maximum So

R

S

Qη 1

Q

So the efficiency of Otto cycle is maximum and the efficiency of diesel cycle will be minimum in this case.

otto Dual Deisel

2) For Same compression Ratio and

heat Rejection Otto cycle 1-2-3-4, Diesel cycle 1-2-3’-4 and Dual cycle 1-2-3”-4”-4 are shown on P-V and T-S diagram. From the T-S diagram it is clear that heat rejection i.e. area of T-S curve in process 4-1 along S- axis is same.

V


But the heat supplied in Otto cycle is maximum and in Diesel cycle is minimum So

S Deisel S ottoQ Qsdual Q

and R

S

Qη 1

Q

So the efficiency of Otto cycle is maximum and the efficiency of diesel cycle will be minimum in this case.

otto Dual Deisel

Examples

Q.1 A Diesel engine has a compression ratio of 20 and cut off takes place at 5% of stroke find air standard efficiency γ 1.4

Solution:

1

1 2 2

2

Vr 20,V 20V 20V

V

S 2 CV 19V 19V

3 S CV 0.05XV V

C C0.05X19V V

3 CV 1.95V

3 C

2 C

V 1.95Vρ 1.95

V V

γ

C

γ 1

C

ρ 11η 1

γ ρ 1r

1.4

4

1 1.95 11 64.9%

1.4 1.95 120

Q.2 4-Cylinder, 2-stroke IC engine has

the following particulars: engine speed = 3000 rpm, bore = 120 mm, crank radius = 60 mm, mechanical efficiency = 90% and the engine develops 75 BHP. Calculate the swept volume and mean effective pressure (MEP).

Solution: Mechanical efficiency,

BrakePower bhp

Indicated power Ihp

750.9

P , So, P = 83.33 hp

We know that Engine Power, P = T ω

Here, P = 83.33 HP = 83.33 x 746 W = 62166.67 W

2πN 2 π 3000

60 60

= 314.16 rad/s T =P/ ω = 62166.67/314.16 = 197.88 N.m

Mean Effective Pressure (MEPorPmep)

C

d

2πN TMEP

V

Vs = K.(π/4).d2. L K = No. of the cylinder Stroke L = 2 x crank radius

= 2 x 0.06 m = 0.12 m

2 3 3Vd 4 0.12 0.12 5.43 10 m4

Nc = No. of cycle per power stroke = 1 for a 2-stroke engine Therefore,


2π 1 197.88

0.ME

0 5P

0 43

=228971.77 N/m2 = 228.97 KPa

Q.3 Two engines are to operate on Otto

and Diesel cycles with the following data: Maximum temperature 1400 K, exhaust temperature 700 K. State of air at the beginning of compression 0.1 MPa, 300 K. Estimate the compression ratios, the maximum pressures, efficiencies, and rate of work outputs (for 1 kg/min of air) of the respective cycles.

Solution:

T3 = 1400 K T4 = 700 K P1= 100 KPa

T1 = 300 K

11

1

RT 0.287 300v

P 100

= 0.861 m3/kg γ 1

γ 1γ

3 3 4

4 4 3

T P V

T P V

γ 1

1

2

V14002

700 V

γ 1

2 1

1 2

T V

V2

T

2T 2 300 600K

1

1 γ 1

2

Vrc 2 5.657

V

Work done W= Q1–Q2=Cv (T3–T2)– Cv (T4 – T1)

=0.718[(1400–600)–(700–300)]KJ/Kg

= 287.2 KJ/Kg. Q.4 A six cylinder ,four stroke spark

ignition engine of 10cm 12cm

(bore stroke) with a compression ratio of 6 is tested at 4800 rpm-on a dynamometer of arm 55 cm. during

a 10 minutes test, the dynamometer reads 45kg and engine consumed 5 kg of petrol of CV 45 MJ/kg. The carburettor receives air at 29 0 C and 1 bar at the rate of 10kg/min. Calculate:- 1) Brake power 2) Brake mean effective pressure 3) Brake specific fuel consumption 4) Brake specific air consumption 5) Brake thermal efficiency 6) Air full ratio

Solution:

6 cylinders, 4 strokes Bore =10cm =0.1m Stroke=0.12m n=6, N=4800rpm Arm of dynamometer=55cm= 0.55m Dynamometer reads 45 Kg in 10min

1. Brake power B.P. = T×ω = 2πNT/60 T= F r = 45× 9.81 × 0.55

= 242.79 Nm. B.P. = 2×3.14×4800×242.79/60 = 122.039KW

2. Brake mean effective pressure

Brake power = pm l a N

n60 2

m

122.039 10 10 10 60 2P

0.12 0.1 0.1 4800 64

π

5.39bar 3. Brake specific fuel consumption.

mB

fsf

.c

B.P

5

601

m0

f

= 30kg/hr

30

122.039

= 0.245kg/KWh 4. Brake specific air consumption

ma

bsacbp


a

kgm =10×60

h

600

=122.039

4.92kg / KWh

5. Brake thermal efficiency

bth

brake powerη

m C.V.

122.039

5 45 10 10 10

10 6

2.54%

0

3

6. Air fuel ratio

A 10

20%F 0.5

Q.5 A sharp edged circular orifice of

diameter 3.8 cm and co-efficient of discharge as 0.6 is used to measure air consumption of a four stroke petrol engine. The pressure drop through the orifice is 145 cm of water and barometer reads 75.5cm of Hg. The compression ratio of the engine is 6 and the piston displacement volume is 2000cm3. The temp of air is taken to be 260C at 2600 rpm, the engine brake power recorded is 29.5 KW. The fuel consumption is 0.14 kg/min and the calorific value of fuel used is 43960 KJ/kg. Calculate following: (1) Volumetric efficiency (2) Air – fuel radio (3) Brake mean effective pressure (4) Brake thermal efficiency

Solution: Diameter of orifice d =3.8 cm

Coefficient of discharge=0.6

Pressure drop =145 mm of water

= 1,422.45 Pa

Barometer reading =75.5 cm of hg

= 1,00,729.08 Pa

r = 6, Vs =2000 cm3

Temperature of air =299k

B.P.=29.5 kw at 2600 rpm

F.C.=0.14 kg/min

CV=43960 kj/kg

1) volumetric efficiency

altual volume innaled / cyclev

sweet volume.

5

31.00729 10Pair 1.1738kg / m

287 299

Now in orifice air enter per second,

Cd 2 Psair

2π0.6 0.038 (2 1422 1.173)

4

= 0.03932 kg/s Sweet volume

6 6 2600

2 60 2000 10

= 0.0433 m3/rev Actual mass enter/sec = 0.03932 kg/s Actual volume enter/sec

0.03932

1.1738

= 0.3349 m3/s

v

0.3349η

0.0 77

436%

3.3

2) air fuel ratio ma= 0.03932 kg/s

Sfc= 0.14 kg/min0.14

kg / s60

=2.33×10-3 kg/s

A 0.0393216.85

F 2.33 163

3) Brake mean effective Pressure

PmB.P.

N

l a

60 2

3

6

60 2 29.5 106.8bar

2600 2000 10

4) brake thermal efficiently

bth

BPη

mf CV

3

3 3

29.5 10

2.33 10 43960 10

28.86%

Q.6 The following data are know for a four cylinder stroke petrol engine: cylinder dimension = 11 cm bore, 13


cm stroke engine speed = 2250 rpm, brake power=50kw, friction power =15 KW

fuel consumption rate =10.5 kg/h, calorific value of fuel= 50,000 KJ/kg,

air inhalation rate = 300 kg/h, ambient condition =150C, 1.03 bar. Estimate:

1. Brake mean effective pressure 2. Brake thermal efficiency 3. Mechanical efficiency

Solution: 1. Brake mean effective pressure

PLANK

B.P60 2

B.P. 60 2

PLANK

2

50 60 2

2250 4 π / 4 0.11 0.13

= 5.396 bar

2. Brake thermal efficiency

BTh

brakepowerη

mf CV

50

10.5 / 3600 X5 3

04

009%

0.2

3. Mechanical Efficiency BP = 50 KW FP = 15 KW IP = 50 + 15 = 65KW

Mechanical efficiencyBP 50

IP 65

= 76.92% Q.7 A liquid fuel C7 h16 is burned with

10% more air than the stoichiometric air assuming complete combustion calculate: 1. The mass of air supplied per kg

of fuel and 2. The volumetric analysis of the

dry products of combustion. Assume air contains 21 % O2 by volume.

Solution:

Equation for stoichiometric composition:

7 16 2 2 2C h 11O 7CO 8H O

If 10% more air is supplied

7 16 2 2

2 2 2 2

79C H 12.1O [12.1x( )]N

21

7CO 8H O 1.1O 45.52N

Mass of O2 for 100 kg of fuel = 12.1 x 32 = 387.3 kg of O2

Mass of air for 100 kg of fuel = 1683.4 kg of air Mass of air required for 1 kg of fuel

=16.82 kg Dry exhaust gas Analysis on volume basis

Moles % Co2

O2

H2O N2

7 1.1 8 45.52

11.36 % 1.78 % 12.9 % 73.87 %

Q.8 Find the percentage increase in the

efficiency of a Diesel engine having a compression ratio of 16 and cut-off ratio (r) is 10% of the swept volume if CV by 2% take Cv = o.717 KJ/Kg.K and γ=1,4

Solution: The efficiency of Diesel engine

γ

γ 1

1 [ρ 1]η 1

r [γ(ρ 1)]

Where

v1

r 16v2

ρ = cut of ratio v3 cp

,v2 cv

Cv = 0.717 KJ/kg k, γ = 1.4 Cp = γ× Cv = 1.0038 kj/kg k R = Cp-Cv = 1.0038 – 0.717 = 0.2868 kj/kg k V3 – V2 = 0.1(V1 – V2) dividing the equation by V2, we get


0.10V3 V1

1 1V2 V2

(ρ-1)=0.10(r - 1) (ρ- 1)=0.10(16 - 1) ρ=2.5

γ

γ 1

1 [ρ 1]η 1

r [γ(ρ 1)]

1.4

1.4 1

1 [2.5 1]1

16 [1.4 (2.5 1)]

= 0.5905 = 59.05 % When cv is decreases by 2% CV’ = 0.70266 R = 0.2868 = cp’-cv’ Cp’ = 0.98946

Cp'

Cv'’ 1.4081

γ

γ' 1

1 [ρ 1]η' 1

r [γ'(ρ 1)]

1.408

1.408 1

1 [2.5 1]1

16 [1.408 (2.5 1)]

= 0.5975 = 59.75 % Increase in efficiency of diesel cycle

η' η 0.5975 0.5905

100η 0.5905

= 1.185 %


Topics

1. THERMODYNAMIC SYSTEM AND PROCESSES

2. FIRST LAW, HEAT, WORK AND ENERGY

3. SECOND LAW, CRANOT CYCLE AND ENROPY

4. AVALILABILITY AND IRREVERSIBILITY

5. PURE SUBSTANCES

6. POWER SYSTEM (RANKINE, BRAYTON, ETC.)

7. IC ENGINE

8. REFRIGERATION AND AIR CONDITIONING

GATE QUESTIONS


Q.1 The following four figures have been drawn to represent a fictitious thermodynamic cycle, on the p -v and T -s planes.

According to the first law of thermodynamics, equal areas are enclosed by a) Figures 1 and 2b) Figures 1 and 3c) Figures 1 and 4d) Figures 2 and 3

[GATE–2005]

Q.2 A reversible thermodynamic cycle containing only three processes and producing work is to be constructed. The constraints are i) There must be one isothermal

process, ii) There must be one isentropic

process, iii) The maximum and minimum

cycle pressures and the clearance volume are fixed, and

iv) Polytropic processes are notallowed.

Then the number of possible cycles are a) 1 b) 2c) 3 d) 4

[GATE–2005]

Q.3 Match items from groups I, II, III, IV and V.

a) F-G-J-K-M b) E-G-I-K-ME-G-I-K-N F-H-I-K-N

c) F-H-J-L-N d) E-G-J-K-NE-H-I-L-M F-H-J-K-M

[GATE–2006]

Common Data for Q.4 and Q.5 A thermodynamic cycle with an ideal gas as working fluid is shown below.

Q.4 The above cycle is represented on T -s plane by a) b)

c) d)

[GATE–2007]

1 THERMODYNAMIC SYSTEM AND PROCESSES


Q.5 If the specific heats of the working fluid are constant and the value of specific heat ratio is 1.4, the thermal efficiency (%) of the cycle is

a) 21 b) 40.9 c) 42.6 d) 59.7

[GATE–2007] Q.6 If a closed system is undergoing an

irreversible process, the entropy of the system

a) must increase b) always remains constant c) Must decrease d) can increase, decrease of remain

constant [ME-GATE –2009]

Q.7 Heat and work are a) intensive properties b) extensive properties c) point functions d) path functions

[ME-GATE –2012] Q.8 A certain amount of an ideal gas is

initially at a pressure p1 and temperature T1. First, it undergoes a constant pressure process 1-2 such that T2 = 3T1/4. Then, it undergoes a constant volume process 2-3 such that T3 =T1/2. The ratio of the final volume to the initial volume of the ideal gas is a) 0.25 b) 0.75 c ) 1.0 d) 1.5

[ME-GATE -2014(3)] Q.9 Two identical metal blocks L and M

(specific heat = 0.4kJ/ kg. K), each having a mass of 5kg, are initially at 313K. A reversible refrigerator extracts heat from block L and rejects heat to block M until the temperature of block L reaches 293K. The final temperature (in K) of block M is _____.

[ME-GATE -2014(4)]

Q.10 Which of the following statements are TRUE with respect to heat and work? i) They are boundary phenomena ii) They are exact differentials iii) They are path functions a) both (i) and (ii) b) both (i) and (iii) c) both (ii) and (iii) d) only (iii)

[GATE -2016(1)] Q.11 A mass m of a perfect gas at pressure p1

and volume V1 undergoes an isothermal process. The final pressure is p2 and volume is V2. The work done on the system is considered positive. If R is the gas constant and T is the temperature, then the work done in the process is

a) 21 1

1

Vp V lnV

b) 11 1

2

pp V lnp

−

c) 2

1

VRT lnV

d) 2

1

pmRT lnp

−

[ME-GATE -2017(1)] Q.12. The volume and temperature of air

(assumed to be an ideal gas) in a closed vessel is 2.87 m3 and 300K, respectively. The gauge pressure indicated by a manometer fitted to the wall of the vessel is 0.5bar. If the gas constant of air is R = 287 J/kg. K and the atmospheric pressure is 1 bar, the mass of air (in kg) in the vessel is

a) 1.67 b) 3.33 c) 5. d) 6.66 [ME-GATE -2018(2)]

Q.13 A n engine operates on the reversible cycle as shown in the figure. The work output from the engine (in kJ/cycle) is ______ (correct to two decimal places).

[ME-GATE -2018(2)]


Q.14 Air is held inside a non-insulated

cylinder using a piston (mass M=25 kg and area A=100 cm2) and stoppers (of negligible area), as shown in the figure. The initial pressure iP and temperature iT of air inside the cylinder are 200 kPa and 400 0 C , respectively. The ambient pressure P∞ and temperature T∞ are 100 kPa and 27 0 C , respectively. The temperature of the air inside the cylinder ( 0 C ) at which the piston will begin to move is ______ (correct to two decimal places).

[ME-GATE-2018 (2)]


1 2 3 4 5 6 7 8 9 10 11 12 13 (a) (d) (d) (c) (a) (d) (d) (b)

(b) (b) (c) 62.5

14 146

ANSWER KEY:


Q.1 (a) From the first law of thermodynamics for a cyclic process ∆U = 0 And δQ δ W=∮ ∮ The symbol ∮ δQ , which is called the cyclic integral of the heat transfer represents the heat transfer during the cycle and ∮ δ W the cyclic integral of the work represents the work during the cycle. We easily see that figure 1 and 2 satisfy the first law of thermo-dynamics, both the figure are in same direction (clockwise) and satisfies the relation. ∮ δQ =∮ δ W

Q.2 (d) Two cycles having constant volume process and two cycles having constant pressure process can be formed. Therefore a total of four cycles can be formed.

Q.3 (d)

Q.4 (c) In the given p - v diagram, three processes are occurred. i) Constant pressure (Process 1 – 2)ii) Constant Volume (Process 2 – 3)iii) Adiabatic (Process 3 – 1)We know that, Constant pressure and constant volume lines are inclined curves in the T-s curve, and adiabatic process is drawn by a vertical line on a T-s curve.

Given P-v diagram is clockwise. So T-s diagram must be clockwise.

Q.5 (a)

This cycle shows the Lenoir cycle. For Lenoir cycle efficiency is given by

1γ

pL

p

r 1η 1 γ

r 1

−

= − −

Where, 2p

1

p 400r 4p 100

= = =

And p

v

Cγ 1.4

C= = (Given)

So, 1

1.4

L(4) 1η 1 1.4 1 0.789

4 1

− = − = − −

= 0.211

Lη 21.1% 21%= ;

Q.6 (d) If a closed system is undergoing an irreversible process, the entropy of the system can increase, decrease or remain constant.

Q.7 (d)

Q. 8 (b) For constant pressure process (1to

2): 1 2

1 2

V VT T

=

EXPLANATIONS


⇒ 2 12 1

1

T 3VV VT 4

= =

For constant volume process (2 to 3)

∴

3 2

1 1

V VFinal volume 3 0.75Initial volume V V 4

= = = =

Q.9 (333 to 335) Q.10 (b) Q.11 Sol: (b)

∵w Pdv= ∫

For ideal gas PV = mRT = const (c )

PV = C ⇒CPV

=

2

1

V

V

CW dvV

= ∫

2

1

VVc[ln v]=

2

1

Vw clnV

=

⇒ 21 1

1

Vw P V lnV

=

When work is done on the system (compression), 2 1V V< and 2 1P P>

For isothermal process

1 1 2 2P V P V c= =

∴ 2 1

1 2

V PV P

=

∴ 11 1

2

Pw P V lnP

=

If the work done on the system is + ive, then

11 1

2

Pw P V lnP

= −

Q.12 (c)

abs gauge atmP P P= +

50 100 150kPa= + =

By ideal gas eqn

absP V mRT=

150 2.87m.287 300

×=

×

m 5kg=

Q.13 (62.5)

For a reversible process,

work output = PdV∫

Hence, for the given reversible cycle, Work Output = Area enclosed by the triangle

1 0.5 250 62.5 kJ/cycle2

= =

Q.14 (146.03)


inside amb

34

22

1 2

1 2

2o

2

mgP PA

25 9.81100 10100 10

P 124.525 kN/mAs mass and volume remains constantP PT T200 124.525673 TT 419.02 K = 146.03 C

−−

= +

= +

=

=

=

=


Q.1 A small steam whistle (perfectly insulated and doing no shaft work) causes a drop of 0.8 kJ/kg in the enthalpy of steam from entry to exit. If the kinetic energy of the steam at entry is negligible, the velocity of the steam at exit is a) 4 m/s b) 40 m/sc) 80 m/s d) 120 m/s

[GATE–2001]

Q.2 A 2 kW, 40 liters water heater is switched on for 20 minutes. The heat capacity Cp for water is 4.2 kJ/kgK. Assuming all the electrical energy has gone into heating the water, increase of the water temperature in degree centigrade is a) 2.7 b) 4.0c) 14.3 d) 25.25

[GATE–2003]

Common Data for Q.3 and 4 Nitrogen gas (molecular weight 28) is enclosed in a cylinder by a piston, at the initial condition of 2 bar, 298 K and 1 m3. In a particular process, the gas slowly expands under isothermal condition, until the volume becomes 2m3. Heat exchange occurs with the atmosphere at 298 K during this process.

Q.3 The work interaction for the Nitrogen gas is a) 200 kJ b) 138.6 kJc) 2 kJ d) -200 kJ

[GATE–2003]

Q.4) The entropy changes for the Universe during the process in kJ/K is a) 0.4652 b) 0.0067c) 0 d) -0.6711

[GATE–2003]

Q.5 A gas contained in a cylinder is compressed, the work required for

compression being 5000 kJ. During the process, heat interaction of 2000 kJ causes the surroundings to be heated. The changes in internal energy of the gas during the process is a) -7000 kJ b) -3000 kJc) +3000 kJ d) +7000 kJ

[GATE–2004]

Common Data For Q.6 and Q.7 A football was inflated to a gauge pressure of 1 bar when the ambient temperature was 15°C. When the game started next day, the air temperature at the stadium was 5°C. Assume that the volume of the football remains constant at 2500 cm3.

Q.6 The amount of heat lost by the air in the football and the gauge pressure of air in the football at the stadium respectively equal a) 30.6 J, 1.94 bar b) 21.8 J, 0.93 barc) 61.1 J, 1.94 bar d) 43.7 J, 0.93 bar

[GATE–2006]

Q.7 Gauge pressure of air to which the ball must have been originally inflated so that it would be equal 1 bar gauge at the stadium is a) 2.23 bar b) 1.94 barc) 1.07 bar d) 1.00 bars

[GATE–2006]

Q.8 Which of the following relationships is valid only for reversible processes undergone by a closed system of simple compressible substance? (neglect changes in kinetic and potential energy) a) Q = dU + W b) Tds = dU+pdvc) Tds = dU + W d) Q = dU + pdv

[GATE–2007]

Q.9 A gas expands in a frictionless piston-cylinder arrangement. The expansion process is very slow, and is resisted by an ambient pressure

2 FIRST LAW, HEAT, WORK AND ENERGY


of 100 kPa. During the expansion process, the pressure of the system (gas) remains constant at 300 kPa. The change in volume of the gas is 0.01m3. The maximum amount of work that could be utilized from the above process is

a) 0 kJ b) 1 kJ c) 2 kJ d) 3 kJ.

[GATE–2008]

Q.10 A balloon containing an ideal gas is initially kept in an evacuated and insulated room. The balloon ruptures and the gas fills up the entire room. Which one of the following statements is TRUE at the end of above process? a) The internal energy of the gas

decreases from its initial value, but the enthalpy remains constant

b) The internal energy of the gas increases from its initial value, but the enthalpy remains constant

c) Both internal energy & enthalpy of the gas remain constant

d) Both internal energy and enthalpy of the gas increase

[GATE–2008]

Q.11 A rigid, insulated tank is initially evacuated. The tank is connected with a supply line through which air (assumed to be ideal gas with constant specific heats) passes at 1MPa, 3500C. A valve connected with the supply line is opened and the tank is charged with air until the final pressure inside the tank reaches 1MPa. The final temperature inside the tank.

a) is greater than 3500C b) is less than 3500C c) is equal to 3500C d) may be greater than, less than, or

equal to, 3500C depending on the volume of the tank

[GATE–2008] Q.12 In a steady state flow process taking

place in a device with a single inlet and a single outlet, the work done per unit mass flow rate is given by

W= −∫ vdpoutletinlet where v is the

specific volume and p is the pressure. The expression for W given above a) is valid only if the process is both

reversible and adiabatic b) is valid only if the process is both

reversible and isothermal c) is valid for any reversible

process d) is incorrect; it must be

outlet

inlet

W pdv= ∫

[GATE–2008] Q.13 A frictionless piston-cylinder device

contains a gas initially at 0.8MPa and 0.015 m3. It expands quasi-statically at constant temperature to a final volume of 0.030 m3. The work output (in kJ) during this process will be

a) 8.32 b) 12.00 c) 554.67 d) 8320.00

[GATE–2009] Q.14 A compressor undergoes a

reversible, steady flow process. The gas at inlet and outlet of the compressor is designated as state 1 and state 2 respectively. Potential and kinetic energy changes are to be ignored. The following notations are used:

v = Specific volume and p = pressure of the gas. The specific work required to be supplied to the


compressor for this gas compression process is

a)2

1

pdv∫ b)2

1

vdp∫

c) ( )1 2 1v p -p d) ( )2 1 2p v v− − [GATE–2009]

Common Data for Q.15 and Q.16 The inlet and the outlet conditions of steam for an adiabatic steam turbine are as indicated in the figure. The notations are as usually followed.

Q.15 If mass rate of steam through the

turbine is 20 kg/s, the power output of the turbine (in MW) is

a) 12.157 b) 12.941 c) 168.001 d) 168.785

[GATE–2009]

Q.16 Assume the above turbine to be part of a simple Rankine cycle. The density of water at the inlet to the pump is 1000 kg/m3. Ignoring kinetic and potential energy effects, the specific work (in kJ/kg) supplied to the pump is

a) 0.293 b) 0.351 c) 2.930 d) 3.510

[GATE–2009]

Common Data For Q.17 and Q.18 The temperature and pressure of air in a large reservoir are 400 K and 3 bar respectively. A converging-diverging nozzle of exit area 0.005 m2 is fitted to the wall of the reservoir as shown in the figure. The static pressure of air at the exit section for isentropic flow through the nozzle is 50 kPa. The characteristic gas constant and the ratio of specific heats of air are 0.287kJ/kgK and 1.4 respectively

Q.17 The density of air in kg/m3 at the nozzle exit is

a) 0.560 b) 0.600 c) 0.727 d) 0.800

[GATE–2011]

Q.18 The mass flow rate of air through the nozzle in kg/s is

a) 1.30 b) 1.77 c) 1.85 d) 2.06

[GATE–2011]

Q.19 The contents of a well-insulated tank are heated by a resistor of 23 Ω in which 10 A current is flowing. Consider the tank along with its contents as a thermodynamic system. The work done by the system and the heat transfer to the system are positive. The rates of heat (Q), work (W) and change in internal energy (∆U) during the process in kW are

a) Q = 0, W = -2.3, ∆U = +2.3 b) Q = +2.3, W = 0, ∆U = +2.3 c) Q = -2.3, W = 0, ∆U = -2.3 d) Q = 0, W = +2.3, ∆U = -2.3

[GATE–2011]

Common Data for Q.20 and Q.21 Air enters an adiabatic nozzle at 300 kPa, 500 K with the velocity of 10 m/s. It leaves the nozzle at 100 kPa with a velocity of 180 m/s. The inlet area is 80 cm2. The specific heat of air Cp is 1008 J/kgK.

Q.20 The exit temperature of the air is a) 516 K b) 532 K c) 484 K d) 468 K

[GATE–2012]

Q.21 The exit area of the nozzle in cm2 is a) 90.1 b) 56.3 c) 4.4 d) 12.9

[GATE–2012]


Q.22 For an ideal gas with constant values of specific heats, for calculation of the specific enthalpy, a) it is sufficient to know only the

temperature b) both temperature and pressure

are required to be known c) both temperature and volume

are required to be known d) both temperature and mass are

required to be known [GATE-2015 Set-1]

Q.23 A well insulated rigid container of volume 1m3 contains 1.0 kg of an ideal gas [Cp = 1000J/kgK) and Cv = 800J/kgK] at a pressure of 105 Pa. A stirrer is rotated at constant rpm in the container for 1000 rotations and the applied torque is 100 N-m. The final temperature of the gas (in K) is ______.

[GATE-2015 Set-1] Q.24 Work is done on adiabatic system

due to which its velocity changes from 10 m/s to 20 m/s, elevation increases by 20m and temperature increases by 1 K. The mass of the system is 10 kg, Cv=100 J/kgK and gravitational acceleration is 10 m/s2. If there is no change in any other component of the energy of the system, the magnitude of total work done (in kJ) on the system is _____.

[GATE-2015 Set-2] Q.25 Steam enters a turbine at 30 bar,

300℃ (u = 2750 kJ/kg, h = 2993 kJ/kg) and exits the turbine as saturated liquid at 15 kPa (u = 225 kJ/kg, h = 226 kJ/kg). Heat loss to the surrounding is 50 kJ/kg of steam flowing through the turbine. Neglecting changes in kinetic energy and potential energy, the work output of the turbine (in kJ/kg of steam) is________

[GATE-2015 Set-3]

Q.26 A mixture of ideal gases has the

following composition by mass

If the Universal gas constant is 8314 J/mol-K, the characteristic gas constant of the mixture (in J/kg.K) is _______.

[GATE-2015 Set-3] Q.27 An ideal gas undergoes a reversible

process in which the pressure varies linearly with volume. The conditions at the start (subscript 1) and at the end (subscript 2) of the process with usual notation are: 𝑝𝑝1 = 100 kPa, 𝑉𝑉1 = 0.2 m3 and 𝑝𝑝2 = 200 kPa, 𝑉𝑉2 = 0.1 m3 and the gas constant, R = 0.275 kJ/kg-K. The magnitude of the work required for the process (in kJ) is ________

Q.28 The internal energy of an ideal gas is

a function of a) temperature and pressure b) volume and pressure c) entropy and pressure d) temperature only

[GATE-2016 Set-2] Q.29 A piston-cylinder device initially

contains 0.4 m3 of air (to be treated as an ideal gas) at 100 kPa and 80℃. The air is now isothermally compressed to 0.1 m3. The work done during this process is _______ kJ. (Take the sign convention such that work done on the system is negative)

[GATE-2016 Set-2]

Q.30 Steam at an initial enthalpy of 100 kJ/kg and inlet velocity of 100 m/s, enters an insulated horizontal nozzle. It leaves the nozzle at 200 m/s. The exit enthalpy (in kJ/kg) is __________

[GATE-2016 Set-3]


Q.31 The molar specific heat at constant volume of an ideal gas is equal to 2.5 times the universal gas constant (8.314 J/mol.K). When the temperature increases by 100K, the change in molar specific enthalpy is _______________ J/mol

[GATE-2017 Set-1] Q.32 A frictionless circular piston of area

210− 2m and mass 100 kg sinks into a cylindrical container of the same area filled with water of density 1000 3 kg / m as shown in the figure. The container has a hole of area

310− 2m at the bottom that is open to the atmosphere. Assuming there is no leakage from the edges of the piston and considering water to be incompressible, the magnitude of the piston velocity (in m/s) at the instant shown is _____ (correct to three decimal places).

[GATE-2018 Set-2]


Q.1 (b) 2 21 2

1 2C Ch h2 2

+ = +

1C 0= 22

1 2Ch h2

− =

23 2C100.

28 × =

2C 40m / s=

Q.2 (c) Heat Supplied H = Power × Time =2 × 1000 × 20 × 60 =2400 × 1000 J But heat required in increasing the water temperature H= mCp∆T 2400×1000=1000×0.04 ×4.2×1000×∆T ∆T=14.3℃

Q.3 (b) For isothermal process,

Work = 21 1

1

VP V lnV

( ) ( )2 100 2ln11

= × × ×

= 138.63 kJ

Q.4 (a)

Q.5 (c) W = -5000 kJ (Negative sign shows that work is done on the system) Q = - 2000 kJ (Negative sign shows that heat is rejected by the system) From the first law of thermodynamics, ∆ Q = ∆W + ∆ U So ∆U = ∆Q − ∆ W = −2000−(−5000) = 3000 kJ

Q.6 (d) Given P1 = 1 bar (Gauge) (P1) absolute = 1+1.013= 2.013 bar.

T1 =15℃=288K T2=5℃=278K Vol. = Constant = 2500 cm3 at constant vol. (15℃→5℃) Q= mCvdT Q= m×718×(278-288) We need to calculate the mass first, Using PV=mRT

5 62.013 10 2500 10 m 287 288−× × × = × ×m = 6.088 × 10−3 kg

Q = 6.088 × 10−3×718 × (278-288) Q = 43.7 J (-ve sign indicates that heat lost by the air in the ball) T1 = 288

P2 = ? T2 = 278

1 2 3 4 5 6 7 8 9 10 11 12 13 14 (b) (c) (b) (a) (c) (d) (c) (d) (c) (c) (a) (c) (a) (b) 15 16 17 18 19 20 21 22 23 24 25 26 27 28 (a) (c) (c) (d) (a) (c) (d) (a) 4.5 2717 (d) 29 30 31 32 - - 29.09 1.456

ANSWER KEY:

EXPLANATIONS


Vol. = C

2 2

1 1

T PT P

=

⇒ 2278P 2.013288

= ×

=1.93 bar. (absolute) ⇒ absol atm gaugeP P P= + ⇒ gaugeP 0.93bar= Q.7 (c) We know that

Since volume is constant

∴ 1 1

2 2

P TP T

=

∴ P2 = 1 bar (gauge) = 2.013 bars (absolute) Substituting the values, we get P1= 2.085 bars (absolute) = 1.07 bar (gauge)

Q.8 (d) δ q = dU + pdv This equation

holds good for a closed system when only pdv work is present. This is true only for a reversible (quasi-static) process.

Q.9 (c) Given: pa = 100 kPa, ps = 300 kPa,

∆v = 0.01 m3 Net pressure across the piston p = ps − pa = 300− 100 =

200 kPa

Maximum work that can be utilized from the system is given by:

W = p∆v = 200 × 0.01 = 2 kJ Q.10 (c)

We know that enthalpy, Given that room is insulated. So

there is no interaction of Energy

(Heat) between system (room) and surrounding (atmosphere).

It means internal Energy dU = 0 and U = constant.

Now flow work pv must also remain constant thus we may conclude that during free expansion process pv i.e. product of pressure and specific volume change in such a way that their product remains constant.

So, it is a constant internal energy and constant enthalpy process.

Q.11 (a)

Given: p1 = 1 MPa, T1 = 350oC= 623 K

For air 𝛾𝛾 = 1.4 We know that final temperature

(T2) inside the tank is given by, T2 = γ T1 = 1.4 × 623 = 872.2 K =

599.2oC T2 is greater than 350oC Q.12 (c) Q.13 (a)

P1 = 0.8 MPa V1 = 0.015 m3 V2 = 0.030 m3 Isothermal process Work output

21 1

1

vW P V lnv

=

( )( )3 0.0300.8 10 0.015 ln0.015

= ×

=0.8 × 103 × 0.015 × ln(2) = 8.32 kJ

Q.14 (b)


Reversible steady flow process.

State 1 : Inlet State 2: Outlet Potential and Kinetic energy change are to be ignored. v= Specific volume of the gas P= Pressure of the gas Specific work required to be supplied to the compressor for this gas compression process is

w = ∫ vdp21

(Compressor is an open flow system)

Q.15 (a)

Apply steady flow energy equation [SFEE]

2 2

1 21 1 2 2

V Vh gz q h gz w2 2

+ + + = + + +

q = 0; because it is an adiabatic steam turbine

⇒ 2

3 (100)10 9.81 10 02

3200× + + × +

=2

3 (100)10 9.81 6 W2

2600× + + × +

⇒ W=607839.24 J/kg = 607.84kJ/kg Mass flow rate of steam through turbine is 20kg/s. The power output of the turbine is =20×607.84

=12156.78 kW=12.157 MW Q.16 (c) Neglecting Kinetic and potential

energy effects SFEE :

h1 + q = h2 + wT (For Turbine) ⇒ 3200=2600+wT ⇒ wT = 600kJ/kg Specific work supplied to the pump is (Since the cycle works between the pressure limits of 3 MPa and 70 kPa)

Wp = � υdP = υ( P1 − P2)

= 1 2P Pρ−

= ( ) 2

3

3000 70 kN / m1000kg / m−

= 2.930kJ/kg

Q.17 (c)

Density of air in (kg/m3) at nozzle exit is

1 2γ γ1 2

P Pρ ρ

=

Where ρ=density

1γ

22 1

1

Pρ .ρP

=

1/1.4 5

250 3 10ρ300 0.287 1000 400

× = × × ×

32ρ 0.727kg / m=

Q.18 (d) Given:

ρ2 = 0.787 kg/m3, A2 = 0.005m2, V2 =? For isentropic expansion,

V2 = ( )1 22 pC T T−

( )32 1.005 10 400 239.73= × × × − = 567.58 m/s

Mass flow rate at exit m = ρ2 A2 V2 =0.727×0.005×567.78

= 2.06 kg/s


Q.19 (a)

W=−I2R [∵work is done on the system] = −(10)2 × 23 = − 2300W = −2.3kW Q = 0 [∵ system is insulated] From 1st law: - ∆Q = ∆U + ∆W ∆U + ∆W = 0

⇒ ∆U = −∆W = +2.3 kW Q.20 (c)

From energy balance for steady flow system

Ein = Eout

2 2

1 21 2

V Vm h m h2 2

+ = +

...(i)

As Equation (1) become

2 2

1 2p 1 p 2

V VC T C T2 2

+ = +

2 2 2 21 2

2 1p

V V 10 180T T 5002 C 2 1008

− −= + = + × × 16.02 500= − +

= 483.98 ≃ 484 K Q.21 (d) From Mass conservation min = mout

1 1 2 2

1 2

V A V Av v

= ...(i)

where v = specific volume of air = RTP

Therefore Eq. (1) becomes

1 1 1 2 2 2

1 2

P V A P V ART RT

=

1 1 1 22

2 2 1

P V A TAP V T× × ×

=× ×

300 10 80 484100 180 500× × ×

=× ×

A2 = 12.9 cm2 Q.22 (a) Q.23 (1283.4 to 1287.4) Q.24 (4.5)

Using SFEE

( ) ( )2 2

1 21 2 1 2

V VW m z z g h h2 2

= − + − + −

( ) ( )2 210 20W 10 20 10 100 1

2 2

= − + − × + ×−

W = − 4.5 kJ ∴ Work done on the system is 4.5 kJ.

Q.25 (2717) By Steady flow energy equation: W = (h1 − h2) q W= (2993 226)− 50 = 2717 kJ/kg

(Neglecting kinetic and potential energy changes)

Q.26 (274 to 276) Q.27 (14.75 to 15.25)

Given: Pressure varies linearly with volume ∴ p = aV+b --------- (A) at V1 = 0.2m3, p1 = 100kPa ∴100 = 0.2a + b ----- (1) at V2 = 0.1m3, p2 = 200kPa 200 = 0.1 a + b --------- (2) From equation (1) & (2), we get a = −1000 and b = 300 ∴ equation (A) ⇒ p = 300 – 1000V

Work 2

1

V

V

pdV= ∫

[ ]2

1

V

V

300 1000V dV= −∫

2

1

V2

V

V 300V 10002

= −

= 300[0.1 – 0.2] – 500 [0.12 – 0.22] Work = −15kJ (-ve sign implies work is done on the system)


∵ in the question only magnitude is asked. ∴ Work = 15kJ

Q.28 (d) Q.29 (-55.6 to -55.4)

Given:- P1 = 100 kPa V1 = 0.4 m3 V2 = 0.1 m3 Work done in an isothermal process is given by

= 21 1

1

VP V lnV

0.1 1000.

0.4 n4

l = ×

W = 55.45kJ Q.30 (84 to 86)

By SFEE:

2 21 2

1 2V Vh h2 2

+ = + [Neglecting

potential head, q = 0 & Wc.v = 0] 2 2

2 3

100 2001h 1002 10

− = +

h2 = 85kJ/kg

Q.31 (29.09)

Given that vC 2.5R=

p vC C R− =

p vC C R 2.5R R 3.5R= + = + =

For ideal gas, Molar specific enthalpy change ( ( H)∆

pH nC dT∆ =

pHH C dT 3.5RdTn∆

∆ = = =

3.5 8.314 100= × ×

2909.9J / mole= Q.32 (1.456)

Given. Piston mass = 100 kg

Pressure at point (1)

5 21 2

100 10P 10 N/m10−= =

Applying Bernoulli’s equation for points (1) and (2):

( )2 2

1 1 2 21 2

P V P V ...... 12 2w w

Z Zg gγ γ

+ + = + +

where P2 = 0 & Z2 = 0 (datum line)

From continuity equation:

A1 V1 = A2 V2 (discharge from hole= volume swept by the piston per unit time)


2 31 2

2 1

10 V 10 VV 10V .....(2)

− −=

⇒ =

From eq. (1) and (2)

( )225

113

21

1

10VV10 0.510 10 2 10 2 10

99Vor 10.5 =20

V 1.456 m/s

+ + =

⇒ =


Q.1 A cyclic heat engine does 50 kJ of work per cycle. If the efficiency of the heat engine is 75%, the heat rejected per cycle is

a) 216 kJ3

b) 331 kJ3

c) 137 kJ2

d) 66 23

kJ

[GATE–2001]

Q.2 A Carnot cycle is having an efficiency of 0.75. If the temperature of the high temperature reservoir is 7270C, what is the temperature of low temperature reservoir? a) 230C b) -230Cc) 00C d) 2500C

[GATE–2002]

Q.3 A solar collector receiving solar radiation at the rate of 0.6 kW/m2 transforms it to the internal energy of a fluid at an overall efficiency of 50%. The fluid heated to 350 K is used to run a heat engine which rejects heat at 315 K. If the heat engine is to deliver 2.5 kW power, the minimum area of the solar collector required would be a) 83.33 2m b) 16.66 2mc) 39.68 2m d) 79.36 2m

[GATE–2004]

Q.4 A heat transformer is device that transfers a part of the heat, supplied to it at an intermediate temperature, to a high temperature reservoir while rejecting the remaining part to a low temperature heat sink. In such a heat transformer, 100 kJ of heat is supplied at 350 K. The maximum amount of heat in kJ that can be transferred to 400 K, when the rest is rejected to a heat sink at 300 K is

a) 12.50 b) 14.29c) 33.33 d) 57.14

[GATE–2007]

Q.5 A cyclic device operates between three thermal reservoirs, as shown in the figure. Heat is transferred to/from the cycle device. It is assumed that heat transfer between each thermal reservoir and the cyclic device takes place across negligible temperature difference. Interactions between the cyclic device and the respective thermal reservoirs that are shown in the figure are all in the form of heat transfer.

The cyclic device can be a) a reversible heat engineb) a reversible heat pump or a

reversible refrigeratorc) an irreversible heat engined) an irreversible heat pump or an

irreversible refrigerator[GATE–2008]

Q.6 An irreversible heat engine extracts heat from a high temperature source at a rate of 100 kW and rejects heat to a sink at a rate of 50 kW. The entire work output of the heat engine is used to drive a reversible heat pump operating between a set of independent isothermal heat reservoirs at 17°C and 75°C. The rate (in kW) at which the heat pump delivers heat to its high temperature sink is a) 50 b) 250

3 SECOND LAW, CARNOT CYCLE AND ENROPY


c) 300 d) 360 [GATE–2009]

Q.7 One kilogram of water at room temperature is brought into contact with a high temperature thermal reservoir. The entropy change of the universe is a) equal to entropy change of the

reservoir b) equal to entropy change of water c) equal to zero d) always positive

[GATE–2010] Q.8 Consider the following two

processes ; a) A heat source at 1200 K loses

2500 kJ of heat to a sink at 800 K b) A heat source at 800 K loses

2000 kJ of heat to a sink at 500 K Which of the following statements is

true ? a) Process I is more irreversible

than Process II b) Process II is more irreversible

than Process I c) Irreversibility associated in both

the processes are equal d) Both the processes are reversible

[GATE–2010] Common Data for Q.9 and Q.10 In an experimental set up, air flows between two stations P and Q adiabatically. The direction of flow depends on the pressure and temperature conditions maintained at P and Q. The conditions at station P are 150 kPa and 350 K. The temperature at station Q is 300 K. The following are the properties and relations pertaining to air: Specific heat at constant pressure, CP = 1.005 kJ/kgK; Specific heat at constant volume, CV = 0.718 kJ/kgK; Characteristic gas constant, R=0.287 kJ/kgK Enthalpy,

Ph C T=

Internal energy, Vu=C T

Q.9 If the air has to flow from station P to station Q, the maximum possible value of pressure in kPa at station Q is close to

a) 50 b) 87 c) 128 d) 150

[GATE–2011]

Q.10 If the pressure at station Q is 50 kPa, the change in entropy ( )Q PS S− in kJ/kgK is

a) -0.155 b) 0 c) 0.160 d) 0.355

[GATE–2011]

Q.11 An ideal gas of mass m and temperature T1 undergoes a reversible isothermal process from an initial pressure P1 to final pressure P2. The heat loss during the process is Q. The entropy change ∆s of the gas is

a) mRln 2

1

PP

b) mRln 1

2

PP

c) mRln 2

1 1

P QP T

−

d) zero

[GATE–2013] Q.12 Which one of the following pairs of

equations describes an irreversible heat engine?

a) δQδQ>0and <0T∮ ∮

b) δQδQ 0and 0T

< <∮ ∮

c) δQδQ 0and 0T

> >∮ ∮

d) δQδQ 0and 0T

< >∮ ∮

[GATE-2014 (3)]

Q.13 A source at a temperature of 500 K provides 1000 kJ of heat. The of temperature environment is 27℃. The maximum useful work (in kJ)


that can be obtained from the heat source is _______.

[GATE-2014 (3)] Q.14 A reversible heat engine receives 2

kJ of heat from a reservoir at 1000 K and a certain amount of heat from a reservoir at 800 K. It rejects 1 kJ of heat to a reservoir at 400 K. The net work output (in kJ) of the cycle is a) 0.8 b) 1.0 c) 1.4 d) 2.0

[GATE-2014 (1)] Q.15 An amount of 100 kW of heat is

transferred through a wall in steady state. One side of the wall is maintained at 127℃ and the other side at 27℃. The entropy generated (in W/K) due to the heat transfer through the wall is ____.

[GATE-2014 (3)] Q.16 A closed system contains 10 kg of

saturated liquid ammonia at 10℃. Heat addition required to convert the entire liquid into saturated vapour at a constant pressure is 16.2 MJ. If the entropy of the saturated liquid is 0.88 kJ/kg.K, the entropy (in kJ/kg. K) of saturated vapor is ____________

[GATE-2014 Set-4] Q.17 A Carnot engine (CE-1) works

between two temperature reservoirs A and B, where TA = 900 K and TB = 500 K. A second Carnot engine (CE-2) works between temperature

reservoirs B and C, where TC = 300 K. In each cycle of CE-1 and CE-2, all the heat rejected by CE-1 to reservoir B is used by CE-2. For one cycle of operation, if the net Q absorbed by CE-1 from reservoir A is 150 MJ, the net heat rejected to reservoir C by CE-2(in MJ) is ____________.

[GATE-2015 (1)]

Q.18 One kg of air (R=287 J/kgK) undergoes an irreversible process between equilibrium state1 (20℃, 0.9 m3) and equilibrium state2 (20℃, 0.6 m3). The change in entropy s2 – s1 (in J/kg.K) is ________.

[GATE-2015 (2)] Q.19 The heat removal rate from a

refrigerated space and the power input to the compressor are 7.2 kW and 1.8 kW, respectively. The coefficient of performance (COP) of the refrigerator is ______

[GATE-2016 (2)] Q.20 A reversible cycle receives 40 kJ of

heat from one heat source at a temperature of 127℃ and 37 kJ from another heat source at 97℃. The heat rejected (in kJ) to the heat sink at 47℃ is __________

[GATE-2016 (2)] Q.21 A heat pump absorbs 10 kW of heat

from outside environment at 250 K while absorbing 15 kW of work. It delivers the heat to a room that must be kept warm at 300K. The Coefficient of Performance (COP) of the heat pump is ___________.

[GATE-2017 (1)] Q.22 One kg of an ideal gas (gas constant,

R = 400 J/kg.K; specific heat at constant volume,

vc 1000J/kg.K= at 1 bar, and 300 K is contained in a sealed rigid cylinder. During an adiabatic process, 100kJ of work is done on the system by a stirrer. The increase in entropy of the system is _________ J/K.

[GATE-2017 (1)] Q.23 A n ideal gas undergoes a process

from state 11 1T 300 K,p 100 )a( kP= = to state 2

2 2 T 600 K,p 500 )a( kP= = . The


specific heats of the ideal gas are : cp = 1 kJ/kg-K and cv = 0.7 kJ/kg-K. The change in specific entropy of the ideal gas from state 1 to state 2 (in kJ/kg-K) is __________(correct to two decimal places).

[GATE-2018 Set-1]

Q.24 Steam flows through a nozzle at a mass flow rate of =m 0.1kg / s . with a heat loss of 5 kW. The enthalpies at inlet and exit are 2500kJ/kg and 2350 kJ/kg, respectively. Assuming negligible velocity at inlet ( 1C 0≈ ), the velocity ( 2C ) of steam (in m/s) at the nozzle exit is __________ (correct to two decimal places).

[GATE-2018 Set-1]

Q.25 For an ideal gas with constant properties undergoing a quasi-static process, which one of the following represents the change of entropy (

s∆ ) from state 1 to 2?

2 2P

1 1

2 2V P

1 1

2 2P V

1 1

2 1V

1 2

T P) s = C ln R lnT P

T Vb) s = C ln C lnT V

T Pc) s = C ln C lnT P

T Vd) s = C ln R lnT V

∆ −

∆ −

∆ −

∆ +

a

[GATE-2018 Set-2]


1 2 3 4 5 6 7 8 9 10 11 12 13 14 (a) (b) (a) (d) (a) (c) (d) (b) (b) (c) (b) (a) 400 (c) 15 16 17 18 19 20 21 22 23 24 25

83.33 6.6 50 -16.36 4 64 1.67 287 0.21 447 (a)

ANSWER KEY:


Q.1 (a)

S

WηQ

=

S

500.75Q

=

SQ 66.67=

RQ 66.67 50=∴ −=16.67kJ

216 kJ3

=

Q.2 (b) L

H

Tη 1T

= −

L0.75 1 T1000

−=

LT 0.251000

=

LT 250K= LT 23= − ℃

Q.3 (a) Internal energy of fluid after absorbing solar radiation =0.5×600=300W/m2

engine315η 1350

= −

WQ

=

1

25000.1Q

=

1Q 25000W∴ =Let A be the minimum area of collector

1Q A 300∴ = × 25000A300

∴ =

A= 83.33m2

Q.4 (d) Given: T1 = 400K, T2 = 300 K, T = 350 K, Q = 100kJ Q1 → Heat transferred to the source by the transformer Q2 → Heat transferred to the sink by the transformer

Applying energy balance on the system

1 2Q Q Q= +

2 1 1Q Q Q 100 Q= − = − ...(i) Apply Claudius inequality on the system.

1 2 1 2

1 2

Q Q Q QQ 100T T T 350 400 300= + = = +

Substitute the value of 2Q from equation (i)

1 1 1 1Q 100 Q Q Q100 100350 400 300 400 300 300

− = + = + −

1100 100 1 1Q350 300 400 300

= + − Solving the above equation, we get Q1=57.14 kJ Therefore the maximum amount of heat that can be transferred at 400 K is 57.14 kJ

Q.5 (a) A heat engine cycle is a thermodynamic cycle in which there is a net heat transfer from higher temperature to a lower temperature device. So it is a Heat Engine.

EXPLANATIONS


Applying Clausius inequality on the system for checking the reversibility of the cyclic device

dQ 0T

=∮

31 2

1 2 3

QQ Q 0T T T

+ − =

3 3 3100 10 50 10 60 10 0

1000 500 300× × ×

+ − =

100+100–200 = 0 Here, the cyclic integral of is zero.

This implies, it is a reversible heat engine.

Q.6 (c)

We know that coefficient of

performance of a Heat pump for the given system is,

3H.P.

Q(COP)W

=

Since it is a reversible heat pump

HH.P.

H L

T(COP)T T

=−

3Q348348 290 50

=−

3348 50Q 300K

58×

∴ = =

Q.7 (d)

We know that, Entropy of universe is always increases.

universeS 0∆ >

( ) ( )system surroundingS S 0∆ + ∆ >

Q.8 (b) We know from the clauses

Inequality,

If dQT∮ = 0, the cycle is reversible

dQ 0T

<∮ the cycle is irreversible

and possible For case (I)

dQ 2500 2500T 1200 800

= −∮

25 25 1.04112 8

kJ / kg= − = −

For case (II)

dQ 2000 2000T 800 500

= −∮

20 20 1.5kJ / kg8 5

= − = −

Since, dQT∮

for the process II is

more in magnitude than process I. Therefore, process (II) is more

irreversible than process (I) Q.9 (b)

Maximum possible value of pressure has been asked in the question, so we take it as a case of reversible flow. Tds = dh − vdp

2 2

1 1p

T PR lT

CP

ds ln n

−

=

The maximum possible value of pressure at station Q can be found as follows:

Pp

Q Q

P

T PR ln 0

TC ln

P

− =

QP1.005 300 0.287 ln

350 150 ln =

QP 87.43kPa= Q.10 (c)


By using Tds=dh–vdP, after simplifying we get:

Q QQ p P

P P

T P(S S ) C ln R ln

T P

− = −

300 500.287 ln350 150

1.005ln − =

= 0.160 Q.11 (b) Q.12 (a)

Clausius inequality for irreversible

heat engine, dQ 0.T

<∮ Heat content

of irreversible heat engine dQ > 0 Q.13 (400)

Maximum work can be derived by using a reversible engine between the two temperature limits.

max s maxW Q η=

Ls

H

TQ 1T

= −

smax s L

H

QW Q TT

= −

Given, sQ 1000kJ=

LT 273 27 300K= + =

HT 500K=

max1000W 1000 300500

∴ = − ×

= 400 kJ Q.14 (c)

From Clausius inequality dQ 0T

=∮ , for a reversible cycle

Or 31 2

1 2 3

QQ Q 0T T T

+ − =

Or 2Q2 1 01000 800 400

+ − =

Or 2Q 0.4kJ= From 1st law of thermodynamics for a cycle:

1 2 3Q Q W Q+ = + Or 2+0.4 = W+1 Or W = 1.4 kJ

Q. 15 (83.33)

2

2 1 gen1

dQS S ST

− = +∫

gen100 00 S400 300

0 +∴ −=

Or gen1 1 1S kW / K3 4 12

= − =

gen1000 WS 83.33W / K12 K

∴ = = =

Q. 16 (6.6)

3

fgfg

h 16.2 10 57.24 kJ /T 283

KS ×= ==

fg 5.75 27. 4 24s =10

kJ / kgK=

Since specific entropy = entropy

mass )

g fg fs s s= + 5.724 0.88= +

= 6.604 kJ/ kg K Q.17 (50)


L11

H1

T 900 500η 1 0.444T 900

−= − = =

Also, 21

1

Qη 1Q

= −

2 83 MQ .33 J=∴

Also, L22

2

T 300η 1 1 0.4T 500

= − = − =

32

2

Qη 1Q

= −

2 5Q 0 MJ=∴ Q.18 (-116.36)

22 1

1

V 0.6s s R In 287 lnV 0.9

− = =

=−116.368 J/kg/K Q.19 (4)

Lref

i/p

QCOPW

=

ref7.2COP 41.8

= =

Q.20 (64)

By Clausius inequality

δQ 0T

=∮

[For reversible cycle]

40 37 Q 0400 370 320

⇒ + − =

Q = 64kJ

Q.21.

Q = 15 + 10 = 25 kw

desired effectCopwork input

=

25 515 3

= =

Cop = 1.67

Q.22. By first law

Q du wδ = + δ

v f i0 C [T T ] 100= − −

[ ]f1 T 300 100× − =

fT 400k=

Entropy change of an ideal gas is given by

f f2 1 v

i i

T VS S C ln R lnT V

− = +

f iV V=Q

f2 1 v

i

TS S C lnT

∴ − =

4001000ln300

=

2 1S S 287.6J / k− =

Q.23 (0.21)


1 1

2 2

p p v v

p v

2 22 1 p

1 1

Ideal gasState 1 : T 300K, P 100kPaState 2 : T 600K, P 500kPac 1kJ/kg-K, c c R, c 0.7kJ/kg-K,

c c 1 0.7 RR = 0.3 kJ/kg-KChangein specific entropy

T Ps s c ln R lnT P

600 5001 ln 0.3ln 0.21 kj/kg-K300 100

− = =

− = =

= − = =

⇒ − = − =

− = −

= − =

Q.24 (447.213)

( )

( )

( )

( )

21 1 1

2cv2 2 2

cv1

1 2

21 2 2

22 1 2

2 32

2

m 0.1 kg/s, Q 5kW heat lossApplying SFEE

1m c Q2

1 m c w2

c 0 and w 0assume

1m h Q m h m c2

1m c m h h Q210.1 c 10 0.1 2500 2350 52

c 447.21

h gz

h gz

z z

−

= =

+ + + =

+ + +

= =

=

+ = +

⇒ = − +

⇒ = − −

=

..

..

. .

.

.. . .

.. .

3 m/s

Q.25 (a)


Q.1 Considering the relationship Tds = dU + Pdv between the entropy (s), internal energy (u), pressure (P), temperature (T) and volume (v), which of the following statements is correct ? a) It is applicable only for a

reversible processb) For an irreversible process,

Tds > du + Pdvc) It is valid only for an ideal gasd) It is equivalent to Ist law, for a

reversible process[GATE–2003(ME)]

Q.2 A steel billet of 2000 kg mass is to be cooled from 1250 K to 450 K. The heat released during this process is to be used as a source of energy. The ambient temperature is 303 K and specific heat of steel is 0.5 kJ/ kgK. The available energy of this billet is a) 490.44 MJ b) 30.95 MJc) 10.35 MJ d) 0.10 MJ

[GATE–2004(ME)]

Q.3 The pressure, temperature and velocity of air flowing in a pipe are 5 bars, 500 K and 50 m/s, respectively. The specific heats of air at constant pressure and at constant volume are 1.005 kJ/ kg K and 0.718 kJ/ kg K, respectively. Neglect potential energy. If the pressure and temperature of the surrounding are 1 bar and 300 K, respectively, the available energy in kJ/kg of the air stream is a) 170 b) 187c) 191 d) 213

[GATE–2013(ME)]

Q.4 The maximum theoretical work obtainable, when a system interacts

to equilibrium with a reference environment, is called a)Entropy b) Enthalpyc) Exergy d) Rothalpy

[GATE-2014 (ME) Set-1]

Q.5 One side of a wall is maintained at 400 K and the other at 300 K. The rate of heat transfer through the wall is 1000 W and the surrounding temperature is 25℃. Assuming no generation of heat within the wall, the irreversibility (in W) due to heat transfer through the wall is _______

[GATE-2015 (ME) Set-3]

Q.6 One kg of an ideal gas (gas constant R = 287 J/kg.K) undergoes an irreversible process from state-1 (1 bar, 300 K) to state -2 (2 bar, 300 K). The change in specific entropy (s2 – s1) of the gas (in J/kg. K) in the process is ___________

[GATE-2017 Set-2]

Q.7` A calorically perfect gas (specific heat at constant pressure 1000 J/kg.K) enters and leaves a gas turbine with the same velocity. The temperatures of the gas at turbine entry and exit are 1100 K and 400 K. respectively. The power produced is 4.6 MW and heat escapes at the rate of 300 kJ/s through the turbine casing. The mass flow rate of the gas (in kg/s) through the turbine is.

a) 6.14 b) 7.00

c) 7.50 d) 8.00

[GATE-2017 Set-2]

4 AVALILABILITY AND IRREVERSIBILITY


1 2 3 4 5 6 7 (d) (a) (b) (c) 248.23 -198.9 (b)

ANSWER KEY:


Q.1 (d) TdS=du+Pdv This equation holds good for any process reversible or irreversible, undergone by a closed system, since it is a relation among properties which are independent of both.

Q.2 (a) ρQ mC T= ∆

Q=2000×(0.5kJ/kg K)×(1250-450) = 800000 kJ

1

2

T 1250S mCln 2000 0.5 lnT 450

∆ = = × ×

=1021.165 0TA. Q s)E (= − ∆

=490439.67 k =490.44 MJ

Q.3 (b) Available energy

( ) ( )2201

1 0 0 1 0CCh h T s s

2 2

− − − + −

=

….. (i) 1 1

1 0 p0 0

T Ps s C ln RlnT P

− = −

500 50.287 ln3

1.005 ln00 1

−

=

=0.05147 ∴ eq.(i) Available energy =1.005 (500 – 300) 300(0.05147)

23(50) 10

2−+ ×

=186.8 kJ/ kg = 187 kJ/ kg

Q.4 (c) Exergy is also known as available energy and it is maximum theoretical work obtainable, when a system

interacts to equilibrium with dead state or reference environment.

Q.5 (248.23) By Gauy-Studola theorem: 0 uniI T ( s)= ∆

uni1000 1000( s)300 400

∆ = −

= 0.8333 W/K ∴ Irreversibility:-I= (25 + 273) 0.833 =248.23 W

Q.6 (-198.9)

Change in specific entropy of ideal gas

f ff i p

i i

T PS S C ln R lnT P

− = −

f i2S S 287 ln1

− = −

f iS S 198.93J / kg.k− = −

Q.7 (b)

By SFEE

2 21 2

1 1 2 2C Ch z g q h z g w2 2

+ + + = + + +

1 2C C= (given)

1 2z z= (assume)

EXPLANATIONS


p 1 p 2mC T Q mC T W− = +

p 2 1

W QmC [T T ]

+=

−

4600 3001 [1100 400]

+=

× −

4900m 7kg / sec700

= =


Q.1 Which combination of the following statements is correct? P: A gas cools upon expansion only

when its Joule-Thomson coefficient is positive in the temperature range of expansion.

Q: For a system undergoing a process, its entropy remains constant only when the process is reversible.

R: The work done by closed system in an adiabatic is a point function.

S: A liquid expands upon freezing when the slope of its fusion curve on pressure- Temperature diagram is negative.

a) R and S b) P and Qc) Q, R and S d) P, Q and R

[GATE–2007]

Q.2 Water has a critical specific volume of 0.003155m3/kg. A closed and rigid steel tank of volume 0.025m3 contains a mixture of water and steam at 0.1MPa. The mass of the mixture is 10 kg. The tank is now slowly heated. The liquid level inside the tank a) will riseb) will fallc) will remain constantd) may rise or fall depending on the

amount of heat transferred[GATE–2007]

Q.3 2 moles of oxygen are mixed adiabatically with another 2 moles of oxygen in mixing chamber, so that the final total pressure and temperature of the mixture become same as those of the individual constituents at their initial states. The universal gas constant is given as R. The change in entropy due to mixing, per mole of oxygen, is given by

a) -Rln2 b) 0c) Rln2 d) Rln4

[GATE–2008]

Common Data For Q.4, 5 and 6 In the figure shown, the system is a pure substance kept in a piston-cylinder arrangement. The system is initially a two-phase mixture containing 1 kg of liquid and 0.03 kg of vapour at a pressure of 100 kPa. Initially, the piston rests on a set of stops, as shown in the figure. A pressure of 200 kPa is required to exactly balance the weight of the piston and the outside atmospheric pressure. Heat transfer takes place into the system until its volume increases by 50%. Heat transfer to the system occurs in such a manner that the piston, when allowed to move, does so in a very slow (quasi-static/quasi-equilibrium) process. The thermal reservoir from which heat is transferred to the system has a temperature of 400°C. Average temperature of the system boundary can be taken as 175°C. The heat transfer to the system is 1 kJ, during which its entropy increases by 10 J/K.

Specific volume of liquid (vf) and vapour (vg) phases, as well as values of saturation temperatures, are given in the table below.

5 PURE SUBSTANCES


Q.4 At the end of the process, which one of the following situations will be true? a) superheated vapour will be left

in the system b) no vapour will be left in the

system c) a liquid + vapour mixture will be

left in the system d) the mixture will exist at a dry

saturated vapour state [GATE–2008]

Q.5 The work done by the system during the process is

a) 0.1 kJ b) 0.2 kJ c) 0.3 kJ d) 0.4 kJ

[GATE–2008]

Q.6 The net entropy generation (considering the system and the thermal reservoir together) during the process is closest to

a) 7.5 J/K b) 7.7 J/K c) 8.5 J/K d) 10 J/K

[GATE–2008] Q.7 A pure substance at 8 MPa and

400℃ is having a specific internal energy of 2864 kJ/kg and a specific volume of 0.03432 m3/kg. Its specific enthalpy (in kJ/ kg) is ____.

[GATE-2014 (2)] Q.8 1.5 kg of water is in saturated liquid

state at 2bar (Vf=0.001061 m3/kg, uf=504.0 kJ/kg, hf = 505kJ/kg). Heat is added in a constant pressure process till the temperature of water reaches 400℃ (v=1.5493m3/ kg, u=2967.0kJ/kg, h=3277.0 kJ/kg). The heat added (in kJ) in the process is ____________.

[GATE-2014 (1)] Q.9 The Vander Waals equation of state

is ( )2

ap υ b RTυ

+ − =

, where p is

pressure, υ is specific volume, T is temperature and R is characteristic gas constant. The SI unit of a is

a) J

kg.K b)

3mkg

c) 5

2

mkgs

d) Pakg

[GATE-2015 (2)] Q.10 A rigid container of volume 0.5 m3

contains 1.0 kg of water at 120℃ 3 3

f g0.00106 m / kg, 0.8908 m(υ g)ν / k= = The state of water is a) Compressed liquid b) Saturated liquid c) A mixture of saturated liquid and

saturated vapour d) Superheated vapor

[GATE-2015 (3)]

Q.11 For water at 25℃, s

s

dpdT

= 0.189

kPa/K (ps is the saturation pressure in kPa and 𝑇𝑇s is the saturation temperature in K) and the specific volume of dry saturated vapour is 43.38 m3/kg. Assume that the specific volume of liquid is negligible in comparison with that of vapour. Using the Clausius- Clapeyron equation, an estimate of the enthalpy of evaporation of water at 25℃ (in kJ/kg) is __________

[GATE-2016 (1)] Q.12 The INCORRECT statement about

the characteristics of critical point of a pure substance is that a) there is no constant temperature

vaporization process b) it has point of inflection with

zero slope c) the ice directly converts from

solid phase to vapor phase d) saturated liquid and saturated

vapor states are identical [GATE-2016 (3)]

Q.13 Which one of the following statements is correct for a superheated vapour?


(A) Its pressure is less than the saturation pressure at a given temperature.

(B) Its temperature is less than the saturation temperature at a given pressure.

(C) Its volume is less than the volume of the saturated vapour at a given temperature.

(D) Its enthalpy is less than the enthalpy of the saturated vapour at a given pressure.

[GATE-2018 Set-1]

Q.14 A vehicle powered by a spark ignition engine follows air standard Otto cycle ( γ=1.4). The engine generates 70 kW while consuming 10.3 kg/hr of fuel. The calorific value of fuel is 44,000 kJ/kg. The compression ratio is _______ (correct to two decimal places).

[GATE-2018 Set-1]

Q.15 A tank of volume 0.05 m3 contains a mixture of saturated water and saturated steam at. The mass of the liquid present is 8 kg. The entropy (in kJ/kg K) of the mixture is ___________ (correct to two decimal places).

Property data for saturated steam and water are:

0sat

3 3f g

fg f

At 200 C,p 1.5538 MPa

v 0.001157 m / kg,v 0.12736 m / kg

s 4.1014kJ / kg K, s 2.3309kJ / kg K

=

= =

= =

[GATE-2018 Set-1]


Q.1 (a)

Q.2 (a) Given

3criυ 0.003155m / kg,=

3V 0.025m ,P 0.1MPa= = and M=10 kg. We know, Rigid means volume is constant. Specific volume,

3s

V 0.025υ 0.0025m / kgm 10

= = =

We see that the critical specific volume is more than the specific volume and during the heating process, both the temperature and the pressure remain constant, but the specific volume increases to the critical volume (i.e. critical point). The critical point is defined as the point at which the saturated liquid and saturated vapour states are identical.

So, point (B) will touch the saturated liquid line and the liquid line will rise the Point O.

Q.3 (b)

Given: 1 2, 1 2T T p p= =Universal Gas constant = R Here given oxygen are mixed adiabatically So, dQ 0=

We know ds 0dQT

= = (since dQ=0)

Q.4 (a) When the vapour is at a temperature greater than the saturation temperature, it is said to exist as superheated vapour. The pressure and temperature of super heated vapour are independent properties, since the temperature may increase while the pressure remains constant. Here vapour is at 400℃ and saturation temperature is 200℃ So, at 200 kPa pressure, superheated vapour will be left in the system.

Q.5 (d) Given 1 2p 100KPa,p 200kPa= =

vl 1kg and m 0. kgm 03 = =

10.03x 0.0291.03

==

1 f 1 fgv v x v= +

= 0.001 + 0.0029(0.1 – 0.001) = 0.0038

1 1 1 V mv 1.03v 0.0039 = = =Now, given that heat transfer takes place into the system until its volume increases by % 50

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 (a) (a) (b) (a) (d) (c) 3138 4158 (c) (c) 2443 (c) (a) 7.61 2.5

ANSWER KEY:

EXPLANATIONS


Therefore, 2 1V 1.5 0 85V .005= = Now Work done 2 2 1P (V V )= −

= 200(0.00585 - 0.0039) =0.4 kJ Q.6 (c)

gen1000S 10673−

+=

= 8.514 J/K Q.7 (3138.56)

h = u + pv h = 2864 + 8 ×103×0.03432 h = 3138.56 kJ/kg

Q.8 (4158) Q = m (h2−h1)

=1.5(3277−505) = 4158 kJ

Q.9 (c) Q.10 (c)

3 30.5v m / kg 0.5m / kg1

= =

Since f gv v v< < the state of water is mixture of saturated water and saturated vapour.

Q.11 (2443.24)

By Causius-Clapeyron equation g f fgs

s g sat g

s s hdPdT v T v

−= =

×

fgh 0.189 298 43.38∴ = × ×

fgh 2443.24kJ / kg= Q.12 (c) Q.13 (a)

sat 1 1

1

1 sat 1

P @ T saturation pressure at T temperatureP pressure of superheated vapour at state 1.P P @ T

→

→

< Q.14 (7.61)

( )

( )

( )

f

f

otto 1

1.4 1

0.4

1.4B.P. = 70kN

m 10.3kg/hrCV = 44000 kJ/kg

. . 70= 100 %10.3m CV 440003600

0.55611 0.556

11 0.556

1 0.4439

r = 7.61

B P

r

r

r

γ

γ

η

η

η −

−

=

=

=

=

= − =

⇒ − =

=

.

.

Q.15 (2.49)


f f3 3

3

3

g

s

f s

Volume of liquid = m V8 0.001157 9.256 100.009256 m

So volume of steam = 0.05 0.0092560.040744m

Volume of steamMass of steam =v

0.0407440.12736

0.319912 kg = mTotal mass of mixture = m m

8 0.319

m−= =

=−

=

=

=

+= +

f s

f fg

9128.319912kg

dryness fraction = x=m m

0.319912 0.038458.319912

So, the entropy of the mixture is given bys = s x s

2.3309 0.03845 4.10142.488kg/kgK

sm=

+

= =

+

= +=


Q.1 The Rateau turbine belongs to the category of` a) pressure compounded turbineb) reaction turbinec) velocity compounded turbined) radial flow turbine

[GATE–2001]

Q.2 The efficiency of superheat Rankine cycle is higher than that of simple Rankine cycle because a) the enthalpy of main stream is

higher for superheat cycleb) the mean temperature of heat

addition is higher for superheatcycle

c) the temperature of steam in thecondenser is high

d) the quality of steam in thecondenser is low.

[GATE–2002]

Q.3 In Rankine cycle, regeneration results in higher efficiency because a) pressure inside the boiler

increasesb) heat is added before steam

enters the low pressure turbinec) average temperature of heat

addition in the boiler increasesd) total work delivered by the

turbine increases [GATE–2003]

Q.4 Considering the variation of static pressure and absolute velocity in an impulse steam turbine, across one row of moving blades a) both pressure and velocity

decreasesb) pressure decreases but velocity

increasesc) pressure remains constant,

while velocity increasesd) pressure remains constant,

while velocity decreases[GATE–2003]

Q.5 In a gas turbine, hot combustion products with the specific heats Cp = 0.98 kJ/kgK, and Cv = 0.7538 kJ/kgK enters the turbine at 20 bar, 1500 K and exits at 1 bar. The isentropic efficiency of the turbine is 0.94. The work developed by the turbine per kg of gas flow is a) 689.64 kJ/kg b) 794.66 kJ/kgc) 1009.72 kJ/kg d) 1312.00 kJ/kg

[GATE–2003]

Q.6 The compression ratio of a gas power plant cycle corresponding to maximum work output for the given temperature limits of Tmin and Tmax will be

a) ( )γ

2 γ 1max

min

TT

−

b) ( )γ

2 γ 1min

max

TT

−

c)

γ 1γ

max

min

TT

−

d)

γ 1γ

min

max

TT

−

[GATE–2004]

Common Data for Q.7 and Q.8: Consider a steam power plant using a reheat cycle as shown. Steam leaves the boiler and enters the turbine at 4 MPa , 3500C (h3 = 3095 kJ/kg). After expansion in the turbine to 400 kPa (h4 = 2609 kJ/kg), the steam is reheated to 350℃ (h5 =3170 kJ/kg) and then expanded in a low pressure turbine to 10 kPa (h6 = 2165 kJ/kg). The specific volume of liquid handled by the pump can be assumed to be

6 POWER SYSTEM (RANKINE, BRAYTON, ETC.)


Q.7 The thermal efficiency of the plant neglecting pump work is

a) 15.8% b) 41.1% c) 48.5% d) 58.6%

[GATE–2004] Q.8 The enthalpy at the pump discharge

(h2) is a) 0.33 kJ/kg b) 3.33 kJ/kg c) 4.0 kJ/k d) 33.3 kJ/kg

[GATE–2004]

Q.9 A p -v diagram has been obtained from a test on a reciprocating compressor. Which of the following represents that diagram?

a)

b)

c)

d)

[GATE–2005]

Q.10 In the velocity diagram shown below, u = blade velocity, C = absolute fluid velocity and W = relative velocity of fluid and the subscripts 1 and 2 refer to inlet and outlet. This diagram is for

a) an impulse turbine b) a reaction turbine c) a centrifugal compressor d) an axial flow compressor

[GATE–2005] Common data for Q.11 and Q.12

In two air standard cycles-one operating in the Otto and the other on the Brayton cycle air is isentropically compressed from 300 to 450K. Heat is added to raise the temperature to 600K in the Otto cycle and to 550K in the Brayton cycle.

Q.11 In ηo and ηB are the efficiencies of the Otto and Bray ton cycles, then a) ηo = 0.25, ηB = 0.18 b) ηo = ηB = 0.33 c) ηo = 0.5, ηB = 0.45 d) it is not possible to calculate the

efficiencies unless the temperature after the expansion is given

[GATE–2005] Q.12 If Wo and WB are work outputs per

unit mass, then a) Wo > WB b) Wo < WB c) Wo = WB d) it is not possible to calculate the

work outputs unless the temperature after the expansion is given

[GATE–2005]


Common Data for Q.13 and Q.14: The following table of properties was printed out for saturated liquid and saturated vapour of ammonia. The titles for only the first two columns are available. All that we know that the other columns (column 3 to 8) contain data on specific properties, namely, internal energy (kJ/kg), enthalpy (kJ/kg) and entropy (kJ/kgK)

Q.13 The specific enthalpy data are in

columns a) 3 and 7 b) 3 and 8 c) 5 and 7 d) 5 and 8

[GATE–2005] Q.14 When saturated liquid at 400C is

throttled to – 20°C, the quality at exit will be

a) 0.189 b) 0.212 c) 0.231 d) 0.788

[GATE–2005] Q.15 Given below is an extract from

steam tables.

Specific enthalpy of water in kJ/kg

at 150 bar and 45°C is a) 203.60 b) 200.53 c) 196.38 d) 188.45

[GATE–2006] Q.16 Determine the correctness or

otherwise Assertion (A) and the Reason (R) Assertion (A): In a power plant working on a Rankin cycle, the regenerative feed water heating improves the efficiency of the steam turbine.

Reason (R): The regenerative feed water heating raises the average temperature of heat addition in the Rankin cycle. a) Both (A) and (R) are true and (R)

is the correct reason for (A) b) Both (A) and (R) are true but (R)

is NOT the correct reason for (A) c) Both (A) and (R) are false d) (A) is false but (R) is true

[GATE–2006] Q.17 Determine the correctness or

otherwise of the following Assertion (A) and the Reason (R).

Assertion (A): Condenser is an essential equipment in a steam power plant.

Reason (R): For the same mass flow rate and the same pressure rise, a water pump requires substantially less power than a steam compressor. a) Both (A) and (R) are true and (R)

is the correct reason for (A) b) Both (A) and (R) are true and (R)

is NOT the correct reason for (A) c) Both (A) and (R) are false d) (A) is false but (R) is true

[GATE–2006]

Q.18 Group I shows different heat addition process in power cycles. Likewise, Group II shows different heat removal processes. Group III lists power cycles. Match items from Groups I, II and III.

Group I Group II Group III P. Pressure Constant

S. Pressure Constant

1.Rankine Cycle

Q. Volume Constant

T. Volume Constant

2. Otto cycle

R. Temperature Constant

U. Temperature Constant

3. Carnot cycle

4. Diesel cycle 5.Brayton cycle

a) P-S-5, R-U-3, P-S-1 , Q-T-2 b) P-S-1, R-U-3, P-S-4, P-T-2

c) R-T-3, P-S-1, P-T-4, Q-S-5 d) P-T-4, R-S-3, P-S-1, P-S-5 [GATE–2006]


Q.19 Which combination of the following statements is correct? The incorporation of reheater in a steam power plant:

P: always increases the thermal efficiency of the plant.

Q: always increases the dryness fraction of steam at condenser inlet

R: always increases the mean temperature of heat addition.

S: always increases the specific work output.

a) P and S b) Q and S c) P, R and S d) P, Q, R and S

[GATE–2007]

Q.20 A thermal power plant operates on a regenerative cycle with a single open feed water heater, as shown in the figure. For the state points shown, the specific enthalpies are: h1 = 2800 kJ/kg and h2 = 200 kJ/kg. The bleed to the feed water heater is 20% of the boiler steam generation rate. The specific enthalpy at state 3 is

a) 720 kJ/kg b) 2280 kJ/kg c) 1500 kJ/kg d) 3000 kJ/kg

[GATE–2008]

Common Data For Q.21 and Q.22 In a steam power plant operating on the Rankine cycle, steam enters the turbine at 4MPa, 350° C and exists at a pressure of 15 kPa. Then it enters the condenser and exits as saturated water. Next, a pump feeds back the water to the boiler. The adiabatic efficiency of the turbine is 90%. The thermodynamic states of water and steam are given in table h is specific enthalpy, s is specific entropy and the specific volume;

subscripts f and g denote saturated liquid state and saturated vapour state.

Q.21 The network output (kJ/kg) of the cycle is

a) 498 b) 775 c) 860 d) 957

[GATE–2010]

Q.22 Heat supplied (kJ/kg) to the cycle is a) 2372 b) 2576

c) 2863 d) 3092 [GATE–2010]

Q.23 The values of enthalpy of steam at the inlet and outlet of a steam turbine in a Rankine cycle are 2800 kJ/kg and 1800 kJ/kg respectively. Neglecting pump work, the specific steam consumption in kg/kWh is

a) 3.60 b) 0.36 c) 0.06 d) 0.01

[GATE–2011]

Q.24 An ideal Brayton cycle, operating between the pressure limits of 1 bar and 6 bar, has minimum and maximum temperature of 300 K and 1500 K. The ratio of specific heats of the working fluid is 1.4. The approximate final temperatures in Kelvin at the end of compression & expansion processes are respectively

a) 500 and 900 b) 900 and 500 c) 500 and 500 d) 900 and 900

[GATE–2011]

Q.25 Specific enthalpy and velocity of steam at inlet and exit of a steam turbine, running under steady state, are as given below:


The rate of heat loss from the turbine per kg of steam flow rate is 5 kW. Neglecting changes in potential energy of steam, the power developed in kW by the steam turbine per kg of steam flow rate is

a) 901.2 b) 911.2 c) 17072.5 d) 17082.5

[GATE–2013]

Common Data For Q.26 and 27 In a simple Brayton cycle, the pressure ratio is 8 and temperatures at the entrance of compressor and turbine are 300 K and 1400 K, respectively. Both compressor and gas turbine have isentropic efficiencies equal to 0.8. For the gas, assume a constant value of Cp (specific heat at constant pressure) equal to 1 kJ/kg-K and ratio of specific heats as 1.4. Neglect changes in kinetic and potential energies.

Q.26 The power required by the compressor in kJ/kg of gas is

a) 194.7 b) 243.4 c) 304.3 d) 378.5

[GATE–2013]

Q.27 The thermal efficiency of the cycle in percentage (%) is

a) 24.8 b) 38.6 c) 44.8 d) 53.1

[GATE–2013]

Q.28 In a power plant, water (density =1000 kg/ m3) is pumped from 80 kPa to 3 MPa. The pump has an isentropic efficiency of 0.85. Assuming that the temperature of the water remains the same, the specific work (in kJ/ kg) supplied to the pump is a) 0.34 b) 2.48 c) 2.92 d) 3.43

[GATE-2014 (1)]

Q.29 The thermal efficiency of an air-standard Brayton cycle in terms of pressure ratio rp and γ (= Cp/Cv) is given by

a) γ 1p

11r −− b) γ

p

11r

−

c) 1/γp

11r

− d) (γ 1)/γp

11r −−

[GATE-2014 (2)]

Q.30 For a gas turbine power plant, identify the correct pair of statements. P. Smaller in size compared to

steam power plant for same power output

Q. Starts quickly compared to steam power plant

R. Works on the principle of Rankine cycle

S. Good compatibility with solid fuel a) P and Q b) R and S c) Q and R d) P and S

[GATE-2014 (3)]

Q.31 An ideal reheat Rankine cycle operates between the pressure limits of 10 kPa and 8 MPa, with reheat being done at 4 MPa. The temperature of steam at the inlets of both turbines is 500℃ and the enthalpy of steam is 3185 kJ/kg at the exit of the high pressure turbine and 2247 kJ/kg at the exit of low pressure turbine. The enthalpy of water at the exit from the pump is 191 kJ/ kg. Use the following table for relevant data

Disregarding the pump work, the cycle efficiency (in percentage) is ____

[GATE-2014 (1)] Q.32 In an ideal Brayton cycle,

atmospheric air (ratio of specific heats,Cp/CV = 1.4, specific heat at constant pressure = 1.005 kJ/kgK) at 1 bar and 300 K is compressed to


8 bar. The maximum temperature in the cycle is limited to 1280 K. If the heat is supplied at the rate of 80 MW, the mass flow rate (in kg/s) of air required in the cycle is

[GATE-2014 (2)]

Q.33 Steam at a velocity of 10m/s enters the impulse turbine stage with symmetrical blading having blade angle 30°. The enthalpy drop in the stage is 100 kJ. The nozzle angle is 20°.The maximum blade efficiency (in percent) is ________.

[GATE-2014 (2)] Q.34 Steam with specific enthalpy (h)

3214 kJ/kg enters an adiabatic turbine operating at steady state with a flow rate 10 kg/s. As it expands, at a point where h is 2920 kJ/kg, 1.5 kg/s is extracted for heating purposes. The remaining 8.5 kg/s further expands to the turbine exit, where h = 2374 kJ/kg. Neglecting changes in kinetic and potential energies, the net power output (in kW) of the turbine is ______.

[GATE-2014 Set-4] Q.35 Which of the following statements

regarding a Rankine cycle with reheating are TRUE? i) increase in average temperature

of heat addition ii) reduction in thermal efficiency iii) drier steam at the turbine exit a) only (i) and (ii) are correct b) only (ii) and (iii) are correct c) only (i) and (iii) are correct d) (i),(ii) and (iii) are correct

[GATE-2015 (2)] Q.36 Steam enters a well insulated

turbine and expands isentropically throughout. At an intermediate pressure, 20 percent of the mass is extracted for process heating and

the remaining steam expands isentropically to 9 kPa Inlet to turbine: P = 14MPa, T = 560℃, h= 3486 kJ/kg, s = 6.6 kJ/kgK Intermediate stage: h = 2776 kJ/kg Exit of turbine: P = 9kPa,

hf =174 kJ/kg, hg =2574 kJ/kg, sf = 0.6kJ/kgK, sg = 8.1kJ/kgK If the flow rate of steam entering the turbine is 100 kg/s, then the work output (in MW) is ____.

[GATE-2015 (1)] Q.37 In a Rankine cycle, the enthalpies at

turbine entry and outlet are 3159 kJ/kg. and 2187 kJ/kg respectively. If the specific pump work is 2 kJ/kg, the specific steam consumption (in kg/kW-h) of the cycle based on the net output is _____.

[GATE-2015 (2)] Q.39 The INCORRECT statement about

regeneration in vapor power cycle is that a) it increases the irreversibility by

adding the liquid with higher energy content to the steam generator

b) heat is exchanged between the expanding fluid in the turbine and the compressed fluid before heat addition

c) the principle is similar to the principle of Stirling gas cycle

d) it is practically implemented by providing feed water heaters

[GATE-2016(1)] Q.40 In a steam power plant operating on

an ideal Rankine cycle, superheated steam enters the turbine at 3 MPa and 350℃. The condenser pressure is 75 kPa. The thermal efficiency of the cycle is ________ percent. Given data: For saturated liquid,


At P=75kPa, hf =384.39kJ/kg, νf =0.001037m3/kg, sf =1.213kJ/kg-K At 75 kPa, hfg = 2278.6 kJ/kg, sfg= 6.2434 kJ/kg-K At P=3MPa & T=350℃ (superheated steam), h = 3115.3 kJ/kg, s = 6.7428 kJ/kg-K

[GATE-2016(1)]

Q.41 Consider a simple gas turbine (Brayton) cycle and a gas turbine cycle with perfect regeneration. In both the cycles, the pressure ratio is 6 and the ratio of the specific heats of the working medium is 1.4. The ratio of minimum to maximum temperatures is 0.3 (with temperatures expressed in K) in the regenerative cycle. The ratio of the thermal efficiency of the simple cycle to that of the regenerative cycle is _________

[GATE-2016(2)] Q.42 In a 3-stage air compressor, the inlet

pressure is 𝑝𝑝1, discharge pressure is 𝑝𝑝4 and the intermediate pressures are 𝑝𝑝2 and 𝑝𝑝3 (p2 < 𝑝𝑝3). The total pressure ratio of the compressor is 10 and the pressure ratios of the stages are equal. If 𝑝𝑝1 = 100 kPa, the value of the pressure 𝑝𝑝3 (in kPa) is __________

[GATE-2016(3)] Q.43 The Pressure ratio across a gas

turbine (for air, specific heat at constant pressure,

pc 1040J/kg.K= and ratio of specific

heats, γ = 1.4 ) is 10. If the inlet temperature to the turbine is 1200K and the isentropic efficiency is 0.9, the gas temperature at turbine exit is ______ K.

[GATE-2017 Set-1]

Q.44 In the Rankine cycle for a steam power plant the turbine entry and exit enthalpies are 2803 kJ/kg and 1800 kJ/kg, respectively. The enthalpies of water at pump entry and exit are 121 kJ/kg and 124 kJ/kg, respectively. The specific steam consumption (in kg/k W.h) of the cycle is ______

[GATE-2017 Set-2]


1 2 3 4 5 6 7 8 9 10 11 12 13 14 (a) (b) (c) (d) (a) (a) (b) (d) (d) (b) (b) (a) (d) (b) 15 16 17 18 19 20 21 22 23 24 25 26 27 28 (d) (a) (b) (a) (b) (a) (c) (c) (a) (a) (a) (c) (a) (d) 29 30 31 32 33 34 35 36 37 38 39 40 41 42 (d) (a) 40.73 108.07 88.3 7581 (c) 125.56 (a) 3.7 26.01 0.8 464.14

43 44

679 3.6

ANSWER KEY:


Q.1 (a) The Rateau turbine is a pressure compounded turbine.

Q.2 (b)

The average temperature at which heat is transferred to steam can be increased without increasing the boiler pressure by superheating the steam to high temperatures. The effect of superheating on the performance of vapour power cycle is shown on a T - s diagram. Thus both the net work and heat input increase as a result of superheating the steam to a higher temperature. The overall effect is an increase in thermal efficiency. Since, the average temperature at which heat is added increases.

Q.3 (c)

The object of the regenerative feed heating cycle is to supply the working fluid to the boiler at same state between 2 and 2’ (rather than at state 2) there by increasing the average temperature of heat addition to the cycle.

Q.4 (d) In impulse turbine, the steam pressure remains constant while it flows through the moving blades, where only the kinetic energy converts into mechanical energy.

Q.5 (a) ∴ Given, T3 = 1500 K Cp = 0.98 kJ/kg K CV = 0.7538 kJ/ kg K.

∴ γ = p

V

C1.3

C=

Now,

γ 1r

4 4

3 3

T PT P

−

=

⇒ 0.31.3

4T 11500 20

=

T4 = 751.37K Now turbine efficiency;

EXPLANATIONS


'

3 4

3 4

T TηT T−

=−

⇒0.94 ='41500 T

1500 751.34−

−

∴ T4′ = 796.25 K Turbine work;

Wt = cp(T3 − T4′ ) = 0.98 (1500 – 796.25) = 689.67 kJ/kg K Q.6 (a)

Work output during the cycle p 3 2 p 4 1mC (T T ) mC (T T )= − − − p 3 4 p 2 1mC (T T ) mC (T T )= − − −

4 2p 3 1

3 1

T TmC T 1 T 1T T

= − − −

Since .

( )1

3 2p

4 1

T T rT T

γ−γ= = =

p1Let x,mC K γ −= = γ

∴ Work/cycle

= 43 1x x

p p

T1K T 1 T 1r r

− − −

For maximum power, differentiate with respect to

⇒ ( )( )x 1

3 1 px 1p p

dW xK T T xr 0dr r

−−

= × − =

(For max.)

⇒ ( ) ( )( )x 131 px 1

p

xT T x rr

−

+=

⇒ 2x 3p

1

TrT

=

∴

12x

3p

1

TrT

=

⇒ ( )2 1

maxp

min

TrT

γγ−

=

Q.7 (b) Power obtained from plant = (h3 − h4) + (h5 − h6) Heat supplied to the plant. = (h3 − h1) − (h5 − h4) Thermal efficiency

= Power obtainedHeat supplied

= ( )( )3095 2609 (3170 2165)3170 2609 (3095 29.3)

− + −− + −

=0.4111=41.11%

Q.8 (d) Enthalpy at exit of pump must be

greater than enthalpy at inlet of pump i.e. h2 must be greater, then h1 =29.3kJ/kg. Among the given four options only one option is greater than h1 =29.3kJ/kg, which is 33.3kJ/kg. Hence option (D) is correct.

Or Pump work = vdP

= 1ρ

(Boiler pressure Condenser

pressure)

= 3

110

(4000 10)=3.99 kJ/kg

h2 =29.3 + 3.99 = 33.3 kJ/kg Q.9 (d)


From above figure, we can easily see that option (d) is same.

Q.10 (b) Velocity of flow is constant

throughout the stage and the diagram is symmetrical hence it is a diagram of reaction turbine.

Q.11 (b) We know that efficiency.

1Otto Brayton

2

Tη η 1T

= = −

Otto Brayton300η η 1450

= = −

61 0.339

= − =

So, Otto Braytonη η 33%= = Q.12 (a)

From the previous part of the

question T3(Otto) = 600 K, T3(Brayton) =

550 K, From the P − v diagram of Otto

cycle, we have W0=Q1–Q2 = CV(T3– T2) − CV (T4–T1) ...(i) For process 3 – 4,

γ 1 γ 1

3 4 14 1 3 2

4 5 2

T V V V V ,V VT V V

− −

= = = =

For process 1 – 2,

γ 1

2 1

1 2

T VT V

−

=

So, 3 2

4 1

T TT T

=

34 1

2

T 600T T 300 400KT 450

= × = × =

And W0 = CV (600 – 450) −CV (400–300) = 150 CV – 100 CV = 50 CV ...(ii) From P − v diagram of Brayton cycle, work done, is,

WB = Q1– Q2=Cp(T3–T2) – Cp (T4 – T1) And

14 3

2

T 300T T 550 366.67KT 450

= × = × =

( ) ( )s pW C 550 450 366.67 300= − − − = 33.33 cp ... (iii)

Dividing equation (ii) by (iii), we get

O v

B p

W 50C 50W 33.33C 33.33γ

= =

p

v

Cγ and γ=1.4

C=

= 50 50 133.33 1.4 46.662

= >×

From this, we see that, WO > WB Q.13 (d) From saturated ammonia table

column 5 and 8 are the specific enthalpy date column.

Q.14 (b) The enthalpy of the fluid before

throttling is equal to the enthalpy of fluid after throttling because in throttling process enthalpy remains constant.

h1 = h2 371.43 = 89.05 + x (1418 – 89.05)

( )f g fh h x h h= + − = 89.05 + x (1328.95)

282.38x 0.2121328.95

= =


Q.15 (d) When the temperature of a liquid is

less than the saturation temperature at the given pressure, the liquid is called compressed liquid (state 2 in figure).The pressure and temperature of compressed liquid may vary independently and a table of properties like the superheated vapour table could be arranged, to give the properties at any p and T. The properties of liquids vary little with pressure. Hence, the properties are taken from the saturation table at the temperature of the compressed liquid. So, from the given table at T = 450 C, Specific enthalpy of water =188.45 kJ/kg.

Q.16 (a) Q.17 (b)

(a) Condenser is essential equipment in a steam power plant because when steam expands in the turbine and leaves the turbine in the form of super saturated steam. It is not economical to feed this steam directly to the boiler. So, condenser is used to condense the steam into water and it is an essential part (equipment) in steam power plant.

Assertion (a) is correct. (b) The compressor and pumps

require power input. The compressor is capable of compressing the gas to very high pressures. Pump work very much like compressor except that they handle liquid instead of gases. Now for same mass flow rate

and the same pressure rise, water

pump require very less power because the specific volume of liquid is very less as compare to specific volume of vapour.

Q.18 (a) Q.19 (b) Whether the cycle efficiency

increases or not depends upon the mean temperature of heat addition.

In practice the use of reheat only gives a small increase in cycle efficiency, but it increases the net work output by making possible the use of higher pressures, keeping the quality of steam at the outlet of the turbine within permissible limits.

Q.20 (a) Given: 2800kJ/kg h2=200kJ/ kg From the given diagram of thermal

power plant, point 1 is directed by the Boiler to the open feed water heater and point 2 is directed by the pump to the open feed water heater. The bleed to the feed water heater is 20% of the boiler steam generation i.e. 20% of h1

So,h2 = 20% of h1+80% of h2 = 0.2×2800 + 0.8× 200 = 720 kJ/kg Q.21 (c) 1h 3092. g5kJ / k= ( )2 f g fh h x h h= + −

Now, ( )1 2 f 2 g2 f 2s s s x s s= = + −


( )6.5821 0.7549 x 8.0085 0.7549= + − x = 0.8033 Dryness fraction at exit of turbine

Hence, h2=225.94+0.8033(2599.1-225.94) h2=2132.3kj/kg Turbine work = h1- h2 =3092.5 – 2132.3 =960.2 kJ/kg

act th isenW W= ×η = 0.9×960.2

= 864.2kJ/kg Pump work =vdP = 0.001014 (4000 – 15) = 4.04079 kJ/kg Net work= Turbine work – Pump work = 864.2 – 4.04 = 860.16 kJ/kg

Q.22 (c) Heat supplied =h1-h4 h4=h3+Pump work =225.94 + 4.04078

= 229.98 kJ/kg Heat supplied = h1-h4

= 3092.5 229.98 = 2862.519 kJ/kg

Q.23 (a) Specific steam consumption

(kg/kWh)

( )3600 3600

Network Done 2900 1900= =

− =3.6kg/kWh Q.24 (a)

P1 = 1bar Tminimum= 300 K P2 = 6 bar Tmaximum = 1500 K

𝛾𝛾 = 1.4

γ 1γ

2 2

1 1

T PT P

−

=

0.41.4

2T 6300 1

=

T2 = 500.5K

0.41.4

3

4

T 6T 1

=

⇒ T4 = 900K Q.25 (a) Applying SFEE

2 2

1 21 2

mV mVmh Q mh W2 2

+ + = + +

⇒2 2180 53250 5 2360 W

2000 2000+ − = + +

(Since m = 1 kg/s) ⇒ W = 901.2 kW

Q.26 (c)

WC = Cp(T2’−T1)

γ 1γ2

p1

T (r )T

−

=

⇒ 0.41.4

2T 300 (8)= × T2 = 543.43K

Efficiency of compressor

2 1c '

2 1

T TηT T

−=

−


0.8= '2 1

543.43 300T T

−−

T2 ′ = 604.28k Wc = 1(604.28− 300) Wc = 304.28 kJ/kg

Power required by the compressor Q.27 (a)

γ 1γ3

p4

T (γ )T

−

=

⇒ 4 0.41.4

1400T 772.86K(8)

= =

Efficiency of turbine

'

3 4T

3 4

T TηT T−

=−

'

4

3 4

1400 T0.8T T

−=

−

0.8 ='

41400 T1400 772.86

−−

T4′ = 898.288 K Work of turbine,

WT = Cp (T3 − T4′) = 1 (1400 – 898.288)

= 501.712 kW/ kg Thermal efficiency of the cycle

T CT

W WηQ−

=

501.712 304.28 24.8%1(1400 604.28)

−= =

−

Q. 28 (d) isenwork vdp=

= ( )2 11 P P

water−

ρ

( )3000 801000

−=

2920 2.92kJ / kg1000

= =

actw 2.92work 3.43

0.85= = =η

Q.29 (d)

Q.30 (a) Q.31 (40.73)

Given, P2=10kPa P1=8kPa P3=4kPa h2s=3185 kJ/kg h4s= 2247 kJ/kg h1=3399 kJ/kg h3=3446 kJ/kg h6s=191 kJ/kg WT = (h1− h2s) + (h3− h4s) QS = (h1− h6s) + (h3− h2s) ∴WT = (3399 − 3185) + (3446 − 2247) =1413 kJ/kg QS = (3399−191) + (3446−3185) =3469 kJ/kg

T

s

W 1413Q 3469

η = =

= 0.4073 Or = 40.73%

Q.32 (108.071)

Given, p

v

C1.4

C= γ =

pC = 1.005 kJ/ kgK P1= 1bar T1= 300K P2= 8bar


T3 = Tmax = 1280 K For process 1-2,

10.4

2 2 1.4

1 1

T P 8T P

γ−γ

= =

= 1.8114 ∴ T2 = 300×1.8114 = 543.43K We know that ( )p 3 2Qs mC T T= − = 80MW (given)

( )6

3

80 10m1.005 10 1280 543.43

×=

× −

= 108.071 kg/s Q.33 (88.3)

We know that for maximum blade efficiency,

cos2α

ρ =

where 1

uV

ρ =

And maximum balde efficiency is given by: ( ) 2

b maxcosη = α

( )2ocos 20 0.883= = = 88.3%

Q.34 (7581)

m1 = 10kg/s m3 = 1.5kg/s m2 = 8.5kg/s h1 = 3214kJ/kg h2 = 2920kJ/kg h3 = 2374 kJ/kg WT = m1 (h1− h2) + m2 (h2 − h3) WT=10(3214−2920)+8.5(2920

2374) WT = 7581kW

Q.35 (c)

Q.36 (125.56)

s1 = s3

⇒ 6.6 = 0.6 + x3 [7.5] x3 = 0.8 h3 = 174 + 0.8 [2574− 174] = 2094 kJ/kg ∴ W = 100[h1 − h2] + 80[h2 − h3] =100[3486-2776]+80[2776- 2094] = 125560 kW = 125.56 MW

Q.37 * Q.38 (3.7)

Specific steam consumption =net

3600W

Now, T 2 1W h h= − = 3159 2187 ∴ TW 972kJ / kg= And pW 2kJ / kg= Thus specific steam consumption

3600972 2

=−

kg/ kWh= 3.7815 kg/ kWh

Q.39 (a) Q.40 (26.01)

For process 1→ 2:- 1 2s s=

6.7428 = 1.213 + 2x (6.2434) 2x 0.8857= ∴ 2h = 384.39 + 0.8857[2278.6]

2h 2408.54kJ / kg=


T 1 2W h h= − = 3115.3 2402.54

TW 712.76kJ / kg= [ ]P f 2 1W v P P= −

= 0.001037 [3000 75] PW 3.033kJ / kg= net T PW W W 709.727kJ / kg= − = 3 1 4Q h h= − Also, p 4 2W h h= − ∴ 4 p 2h W h= + = 3.033+384.3

4h 387.423kJ / kg= ∴ sQ 3115.3 387.423= −

= 2727.87 kJ/ kg

netth

s

WQ

η =

∴ th709.727 0.26012727.87

η = =

th 26.01%η =

Q.41 (0.8)

Efficiency without regeneration:-

( )th 1/

p

11r

γ− γη = −

( )th 0.4/1.411 0.4006

6η = − =

Efficiency with regeneration:-

( ) ( )1

minth preg

max

T1 rT

γ−γη = −

( )0.4/1.41 6 0.3= − × = 0.4994 ≈ 0.5

( )

th

th reg

0.4Ratio0.5

η= =

η

= 0.8

Q.42 (464.14) We know that in multi-stage compression

( ) ( )N

p p0 sr r=

( ) ( )1/Np 0s

r rp=

Where N-Number of stages ∴ ( ) 3

p sr 10=

( )p sr 2.1544=

2

1

p 2.1544p

=

2p 2.1544 100= × = 215.44 kPa

Also s

2

p 2.1544p

=

3p 2.1544 215.44= × ∴

3p 464.14kPa= Q.43 (679.38)

3 4isen

3 4

T TT T '

−η =

−

32

1 4

PP 10P P

= =Q

r 1r

4 4

3 3

T ' PT P

−

=

1.4 11.4

41T ' 1200

10

−

=

4T ' 621.53k=

3 40.9[1200 621.53] T T− = −

4T 679.38k=

Q.44 (3.6)


Given

1h 2803kJ / kg=

2h 1800kJ / kg=

3h 121kJ / kg=

4h 124kJ / kg=

T 1 2W h h 2803 1800= − = −

TW 1003kJ / kg=

p 4 3W h h= −

124 121= −

pW 3kJ / kg=

net T pW W W= −

1003 3= −

netW 1000kJ / kg=

Qnet

3600 kgSSCW kwh

=

3600SSC 3.6kg / kwh1000

= =


Q.1 A single-acting two-stage compressor with complete inter cooling delivers air at 16 bar. Assuming an intake state of 1 bar at 15°C, the pressure ratio per stage is a) 16 b) 8c) 4 d) 2

[GATE–2001]

Q.2 In a spark ignition engine working on the ideal Otto cycle, the compression ratio is 5.5. The work output per cycle (i.e., area of the P-v diagram) is equal to 23.625×105×Vc(in Joules), where Vc is the clearance volume in m3. The indicated mean effective pressure is a) 4.295 bar b) 5.250 barc) 86.870 bar d) 106.300 bar

[GATE–2001]

Q.3 An ideal air standard Otto cycle has a compression ratio of 8.5. If the ratio of the specific heats of air (𝛾𝛾) is 1.4, what is the thermal efficiency in percentage of the Otto cycle ? a) 57.5 b) 45.7c) 52.5 d) 95

[GATE–2002]

Q.4 For a spark ignition engine, the equivalence ratio (∅) of mixture entering the combustion chamber has values a) ∅ < 1 for idling and ∅ > 1 for

peak power conditionsb) ∅> 1 for both idling and peak

power conditionsc) ∅ > 1 for idling and ∅ < 1 for

peak power conditionsd) ∅ < 1 for both idling and peak

power conditions [GATE–2003]

Q.5 A diesel engine is usually more efficient than a spark ignition engine because a) diesel being a heavier

hydrocarbon, releases more heatper kg than gasoline

b) the air standard efficiency ofdiesel cycle is higher than theOtto cycle, at a fixedcompression ratio

c) the compression ratio of a dieselengine is higher than that of anSI engine

d) self ignition temperature ofdiesel is higher than that ofgasoline

[GATE–2003]

Q.6 An automobile engine operates at a fuel air ratio of 0.05, volumetric efficiency of 90% and indicated thermal efficiency of 30%. Given that the calorific value of the fuel is 45 MJ/kg and the density of air at intake is 1 kg/m3, the indicated mean effective pressure for the engine is a) 6.075 bar b) 6.75 barc) 67.5 bar d) 243 bar

[GATE–2003]

Q.7 For an engine operating on air standard Otto cycle, the clearance volume is 10% of the swept volume. The specific heat ratio of air is 1.4. The air standard cycle efficiency is a) 38.3% b) 39.8%c) 60.2% d) 61.7%

[GATE–2003]

Q.8 At the time of starting, idling and low speed operation, the carburetor supplies a mixture which can be termed as

7 IC ENGINE


a) Lean b) slightly leaner than stoichiometric c) stoichiometric d) rich

[GATE–2004] Q.9 During a Morse test on a 4 cylinder

engine, the following measurements of brake power were taken at constant speed.

All cylinders firing 3037 kW Number 1 cylinder not firing 2102

kW Number 2 cylinder not firing 2102



kW The mechanical efficiency of the

engine is a) 91.53% b) 85.07% c) 81.07% d) 61.22%

[GATE–2004] Q.10 An engine working on air standard

Otto cycle has a cylinder diameter of 10 cm and stroke length of 15 cm. The ratio of specific heats for air is 1.4. If the clearance volume is 196.3 cc and the heat supplied per kg of air per cycle is 1800 kJ/kg, the work output per cycle per kg of air is

a) 879.1 kJ b) 890.2 kJ c) 895.3 kJ d) 973.5 kJ

[GATE–2004] Q.11 The stroke and bore of a four stroke

spark ignition engine are 250 mm and 200 mm respectively. The clearance volume is 0.001m3. If the specific heat ratio � = 1.4, the air-standard cycle efficiency of the engine is

a) 46.40% b) 56.10% c) 58.20% d) 62.80%

[GATE–2007]

Q.12 Which one of the following is NOT a necessary assumption for the air-standard Otto cycle? a) All processes are both internally

as well as externally reversible. b) Intake and exhaust processes are

constant volume heat rejection processes.

c) The combustion process is a constant volume heat addition process.

d) The working fluid is an ideal gas with constant specific heats

[GATE–2008] Q.13 In an air-standard Otto-cycle, the

compression ratio is 10. The condition at the beginning of the compression process is 100 kPa and 27°C. Heat added at constant volume is 1500 kJ/kg, while 700 kJ/kg of heat is rejected during the other constant volume process in the cycle. Specific gas constant for air = 0.287 kJ/kg K. The mean effective pressure (in kPa) of the cycle is

a) 103 b) 310 c) 515 d) 1032

[GATE–2009] Q.14 A turbo-charged four-stroke direct

injection diesel engine has a displacement volume of 0.0259 m3 (25.9 litres). The engine has an output of 950 kW at 2200 rpm. The mean effective pressure (in MPa) is closest to

a) 2 b) 1 c) 0.2 d) 0.1

[GATE–2010] Q.15 The crank radius of a single-cylinder

I.C. engine is 60 mm and the diameter of the cylinder is 80 mm. The swept volume of the cylinder in cm3 is

a) 48 b) 96 c) 302 d) 603


[GATE–2011] Q.16 In an air-standard Otto cycle, air is

supplied at 0.1 MPa and 308 K. The ratio of the specific heats (γ) and the specific gas constant (R) of air are 1.4 and 288.8 J/kgK respectively. If the compression ratio is 8 and the maximum temperature in the cycle is 2660 K, the heat (in kJ/kg) supplied to the engine is ________

[GATE-2014 (1)] Q.17 A diesel engine has a compression

ratio of 17 and cut-off takes place at 10% of the stroke. Assuming ratio of specific heats (γ) as 1.4, the air–standard efficiency (in percent) is _____________.

[GATE-2014 (3)] Q.18 In a compression ignition engine,

the inlet air pressure is 1 bar and the pressure at the end of isentropic compression is 32.42 bars. The expansion ratio is 8. Assuming ratio of specific heats (γ) as 1.4, the air standard efficiency (in percent) is _______________.

[GATE-2014 Set-4]

Q.19 Air enters a diesel engine with a density of 1.0 kg/m3. The compression ratio is 21. At steady state, the air intake is 30×10−3kg/s and the net work output is 15 kW. The mean effective pressure (kPa) is ______.

[GATE-2015 (1)] Q.20 An air-standard Diesel cycle consists

of the following processes: 1-2 : Air is compressed isentropically. 2-3 :Heat is added at constant pressure. 3-4 :Air expands isentropically to the original volume. 4-1:Heat is rejected at constant volume.

If 𝛾𝛾 and T denote the specific heat ratio and temperature, respectively, the efficiency of the cycle is

a) 4 1

3 2

T T1T T

−−

− b) 4 1

3 2

T T1γ[T T ]

−−

−

c) 4 1

3 2

γ[T T ]1T T

−−

− d) ( )

( )4 1

3 2

T T1

(γ 1) T T−

−− −

[GATE-2015 (3)]

Q.21 For the same values of peak pressure, peak temperature and heat rejection, the correct order of efficiency for Otto, Dual and Diesel cycles is a) ηOtto > ηDual > ηDiesel b) ηDiesel > ηDual > ηOtto c) ηDual > ηDiesel > ηOtto d) ηDiesel > ηOtto > ηDual

[GATE-2015 (2)] Q22 Air contains 79% N2 and 21% O2 on a

molar basis. Methane (CH4) is burned with 50% excess air than required stoichiometrically. Assuming complete combustion of methane, the molar percentage of N2 in the products is _________

[GATE-2017 Set-1]

Q.23 An engine working on air standard Otto cycle is supplied with air at 0.1 MPa and 035 C . The compression ratio is 8. The heat supplied is 500 kJ/kg. Property data for air: pc 1.005= kJ/kg

K, vc 0.718= kJ/kg K, R = 0.287 kJ/kg K. The maximum temperature (in K) of the cycle is _________ (correct to one decimal place).

[GATE-2018 Set-1]


1 2 3 4 5 6 7 8 9 10 11 12 13 14 (c) (b) (a) (b) (c) (a) (d) (d) (c) (d) (c) (b) (d) (a) 15 16 17 18 19 20 21 22 23 (d) 525 (b) (b) 73.8 1403

ANSWER KEY:


Q.1 (c) Given that the intercooling is perfect ∴ P2 = �P1P3 P2 = √1 × 16 P2 = 4 bar

Pressure ratio per stage 32

1 2

PP 4P P

= = =

Q.2 (b)

mep = s 1 C

W WV V V

=−

= 5

C

C C

23.625 10 V5.5V V

× ×−

= 5

C

C

23.625 10 V4.5V×

=5.25× 105Pa =5.250 bar

Q.3 (a) γ 11η 1

r

− = −

0.4118.5

= −

= 0.5751 or 57.51%

Q.4 (b) Equivalence ratio is defined as the actual fuel air ratio to stoichiometric fuel air ratio

actual

Stoi

FAFA

φ

=

For both idling and peak power conditions the actual fuel-air ratio required is more than the stoichiometric fuel-air ratio.

Q.5 (c)

For same compression ratio and the same heat supplied, Otto cycle is most efficient and diesel cycle is least efficient. In practice, however, the compression ratio of the Diesel engine ranges between 14 and 25 whereas that of the Otto engine between 6 and 12. Because of it the efficiency of Diesel cycle is higher efficiency than that of Otto engine.

Q.6 (a)

Volumetric efficiency = actual volumeswept volume

= a

s

V 0.9V

=

∴ Va = 0.9Vs Mass of air,

a air s sm ρ V 0.9V= =mf = 0.05 × 0.9Vs = 0.045Vs

mepthermal

f

P LANη

m C.V×

=×

⇒ 0.3 = mep s6

s

P V0.045v 45 10

×

× ×∴ mepP 6.075bar=

Q.7 (d) γ 1

cOtto

1

vη 1v

−

= −

γ 1

c

c s

V1V V

−

= − + Where Vc = clearance vol = 10% of Vs=0.1Vs

∴ γ 1

sOtto

s s

0.1Vη 10.1V V

−

= − +

= 1.4 10.1

1.1

1

− −

=0.6167=61.67%

EXPLANATIONS


Q.8 (d) Q.9 (c) When cylinder 1 is not firing then

power is 2102 kW and when all cylinders are firing then power is 3037 kW.

Hence power supplied by cylinder No.1= 3037-2102 = 935 kW

Similarly power supplied by cylinder No. 2 = 3037-2102 = 935 kW

Similarly power supplied by cylinder No. 3 = 3037-2100 = 937 kW

Similarly power supplied by cylinder No.4 = 3037-2098 = 939 kW

totalIP 935 935 937 939 3746= + + + =

Hence mech3037η

935 935 937 939=

+ + +

mechBPηIP

=

= 30373746

= 0.8107

∴ ηmech=81.07% Q.10 (d)

r 1

cOtto

c s

Vη 1V V

−

= − +

VC = 196.3cc

2 2s

π πV D L 10 154 4

= × = × ×

=1178.097cc 𝛾𝛾=1.4

∴ 1.4 1

Otto196.3η 1

196.3 1178.097

− = − +

=0.5408 ∴ ηOtto = 54.08% ∴ Work output = ηOtto × (Heat supplied) = 0.5408 ×1800 =973.58kJ Q.11 (c) Given L=250 mm = 0.25 m, D = 220

m = 0.2 m.

p3c

v

Cv 0.001m , γ 1.4

C= = =

Swept Volume 2

sπV A L (D) L4

= × = ×

2 3π (0.2) 0.25 0.00785m4

= × =

Compression ratio c sr

c c

v vvrv v

+= =

0.001 0.007850.001+

=

= 8.85 Air standard efficiency

γ 1 1.4 1

1 1η 1 1(r) (8.85)− −= − = −

11 1 0.418 0.5822.39

= − = − =

or 58.2 %

Q.12 (b)

Assumptions of air standard otto

cycle a) All processes are internally

reversible. b) Air behaves as ideal gas c) Specific heats remains constant

(Cp & Cv) d) Intake process is constant

volume heat addition process and exhaust pro-cess is constant volume heat rejection process. Intake process is a constant volume heat addition process, from the given options; option (2) is incorrect.

Q.13 (d)


P1 = 100 kPa Compression ratio, r = 10

T1 = 27 + 273 = 300K Heat added,

QS = 1500kJ/kg Heat rejected, QR=700kJ/kg

Mean effective pressure

= Work done per cycleSwept volume

Compression ratio, r = V1/V2 = 10

V1 = 10V2 Swept volume

= V1 − V2

= 11 1

VV 0.9V10

− =

For initial air P1 V1 = RT1

⇒ 11

1

RTVP

=

= 0.287kJ / kgK 300K100kPa

×

= 0.861m3/kg Swept volume = 0.9 V1 = 0.9(0.861) = 0.7749m3/kg Work done in cycle,

Wnet out = Qsupply − Qrej = 1500 − 700 = 800 kJ/kg Mean effective pressure

= netWSwept volume

mep.800P

0.7749=

Pmep. = 1032.39 kPa Q.14 (a)

Power mP LAn kw60,000

=

As four stroke

n = N 1100rpm2=

m950 60000P

0.0259 1100×

=×

= 2 ×106N/m2= 2 MPa Q.15 (d) Given : r = 60 mm, D = 80 mm= 8 cm Stroke length, L = 2r = 2 × 60 = 120 mm =12 cm

Swept Volume, Vs = A × L

= 2 2π πD L (8) 124 4

× = ×

= 603.18 ≃ 603 cm3 Q.16 (1400 to 1420)

Given, P1 = 0.1 MPa T1= 308 K 𝛾𝛾 = 1.4 R = 288.8 J/kg K r = 8 T3 = 2660 K

vR 288.8Cγ 1 0.4

= =−

= 722 J/kgK

( )γ 1

2 1

1 2

0.4T VT V

8−

==

0.42T 308 8 707.6K= × =

Qs = Cv ∆T = 722(2660 − 707.6) = 1409.6 kJ/kg

Q.17 (58 to 62)


1

2

V 17V

=

p

v

Cγ 1.4

C= =

or 3 2 1 2V V 0.1(V V )− = −

or 3 1

2 2

V V1 0.1 1V V

− = −

or 3

2

V 0.1 16 1 2.6V

= × + =

γc

Diesel γ 1c

r 11η 1r γ(r 1)−

−= − −

=1.4

0.4

1 2.6 1117 1.4(2.6 1)

−− −

= 0.4

1 3.81 1117 1.4 1.6

− − ×

= 0.596 or 59.6% Q.18 (59 to 61)

Given P1= 1 bar, P2=32.42bar

γ = p

V

C1.4

C=

4 1

3 3

V V 8V V

= =

for process 1 ⟶2, γ γ

1 1 2 2P V P V=

∴ γ

1 2

2 1

V P 32.42V P

= =

Or 1/1.41

2

V (32.42) 11.999 12V

= = ≈

3 1c

2 1

V V 12r 1.5V 8 V

= = × =

γc

d γ 1c

r 11η 1r γ(r 1)−

−= − −

= 1.4

0.4

1 1.5 1112 1.4 0.5

−− ×

= 0.596 or 59.6 % Q.19 (525) Q.20 (b)

Heat applied,Qs = Cp(T3 − T2) Heat rejected, Qr = Cv(T4 − T1)

r 4 1

s 3 2

Q (T T )11Q γ (T T )

1η = −−

= −−

Q.21 (b) Q.22 (73.83)

100 Ltr air contains 79 ltr of N2 and 21 ltr of O2

The combustion equation with 50% excess air is

4 2 2

2 2 2 2

CH 1.5 2[3.762N O ]2H O CO 3 3.3762N O+ × +

→ + + × +

It is assumed that nitrogen is insert and does not participate in the reaction. Whatever nitrogen is there in reactants, same will come out in the products.

On the basis of Molar or volume


23 3.762N 100

2 1 3 3.762 1×

= ×+ + × +

2N 73.83%=

Q.23 (1403.9)

o1 1

1

2

s

p

v

3 max

P 0.1 MPa, T 35 C = 308KV n 8VQ 500 kJ/kgc 1.005kJ/kgK

c 0.718 kJ/kgKR = 0.287 kJ/kgK

1.399 1.40

T T ?For process1 2

p

v

cc

γ

= =

= =

=

=

=

= =

= =−

( )

( )

1 1 2 2

1.412 1

2

2

1 1 2 2 2 22 1

1 2 1 1

2

2

s v 3 2

For process 1 2P V P V

VP P 0.1 8V

P 1.8379 MPaP V P V P Vand T TT T P V

1.8379 1T 3080.1 8

T 707.6KFor process 2 3

Q C T T 500 kJ/kg

−

=

= =

=

= ⇒ =

=

=→

= − =

γ γ

γ

( )3

3

0.718 707.6 500

T 1403.97 K

T − =

=


Q.1 An industrial heat pump operates between the temperatures of 027 C and

013 C− . The rates of heat addition and heat rejection are 750 W and 1000 W, respectively. The COP for the heat pump is a) 7.5 b) 6.5c) 4.0 d) 3.0

[GATE-03]

Q.2 For air with a relative humidity of 80% a) the dry bulb temperature is less thanthe wet bulb temperature b) the dew point temperature is lessthan wet bulb temperature c) the dew point and wet bulbtemperature are equal d) the dry bulb and dew pointtemperature are equal

Common Data For Q.3 and Q.4 A refrigerator based on ideal vapour compression cycle operates between the temperature limits of 020 C− and 040 C . The refrigerant enters the condenser as saturated vapour and leaves as saturated liquid. The enthalpy and entropy values for saturated liquid and vapour at these temperatures are given in the table below.

Q.3 If refrigerant circulation rate is 0.025 kg/s, the refrigeration effect is equal to a) 2.1 kW b) 2.5 kWc) 3.0 kW d) 4.0 kW

[GATE-03]

Q.4 The COP of the refrigerator is a) 2.0 b) 2.33c) 5.0 d) 6.0

[GATE-03]

Q.5 In the window air conditioner, the expansion device used is a) capillary tubeb) thermostatic expansion valvec) automatic expansion valved) float valve

[GATE-04]

Q.6 During the chemical dehumidification process of air a) dry bulb temperature and specifichumidity decreases b) dry bulb temperature increases andspecific humidity decreases c) dry bulb temperature decreases andspecific humidity increases d) dry bulb temperature and specifichumidity increases

[GATE-04]

Q.7 Environment friendly refrigerant R134 is used in the new generation domestic refrigerators. Its chemical formula is a) CHClF2 b) C2Cl3F3c) C2Cl2F4 d) C2H2F4

[GATE-04]

Q.8 A heat engine having an efficiency of 70% is used to drive a refrigerator having a coefficient of performance of 5. The energy absorbed from low temperature reservoir by the refrigerator for each kJ of energy absorbed from high temperature source by the engine is a) 0.14 kJ b) 0.71 kJ

7 IC ENGINE


c) 3.5 kJ d) 7.1 kJ

[GATE-04] Q.9 Dew point temperature of air at one atmospheric pressure (1.013 bar) is

018 C . The air dry bulb temperature is 030 C . The saturation pressure of water at

018 C and 030 C are 0.02062 bar and 0.04241 bar respectively. The specific heat of air and water vapour respectively are 1.005 and 1.88 kJ/kg K and the latent heat of vaporization of water at 00 C is 2500 kJ/kg. The specific humidity (kg/kg of dry air) and enthalpy (kJ/kg or dry air) of this moist air respectively, are a) 0.01051, 52.64 b) 0.01291, 63.15 c) 0.01481, 78.60 d) 0.01532, 81.40

[GATE-04] Q.10 A R-12 refrigerant reciprocating compressor operates between the condensing temperature of 030 C and evaporator temperature of 020 C− . The clearance volume ratio of the compressor is 0.03. Specific heat ratio of the vapour is 1.15 and the specific volume at the suction is 0.1089 m3/kg. Other properties at various states are given in the figure. To realize 2 tons of refrigeration, the actual volume displacement rate considering the effect of clearance is

a) 3 36.35 10 m / s−× b) 3 363.5 10 m / s−× c) 3 3635 10 m / s−× d) 3 34.88 10 m / s−×

[GATE-04] Q.11 For a typical sample of ambient air (at 035 C , 75% relative humidity and standard atmosphere pressure), the amount of moisture in kg per kg of dry air will be approximately a) 0.002 b) 0.027 c) 0.25 d) 0.75

[GATE-05] Q.12 Water at 042 C is sprayed into a stream of air at atmospheric pressure, dry bulb temperature of 040 C and a wet bulb temperature of 020 C . The air leaving the spray humidifier is not saturated. Which of the following statements is true ? a) Air gets cooled and humidified b) Air gets heated and humidified c) Air gets heated and dehumidified d) Air gets cooled and dehumidified

[GATE-05] Q.13 The vapour compression refrigeration cycle is represented as shown in the figure below, with state 1 being the exit of the evaporator. The coordinate system used in this figure is


a) p-h b) T -s c) p-s d) T –h

[GATE-05] Q.14 Various psychometric processes are shown in the figure below

The matching pairs are a) P-(i), Q-(ii), R-(iii), S-(iv), T-(v) b) P-(ii), Q-(i), R-(iii), S-(v), T-(iv) c) P-(ii), Q-(i), R-(iii), S-(iv), T-(v) d) P-(iii), Q-(iv), R-(v), S-(i), T-(ii)

[GATE-05]

Q.15 A vapour absorption refrigeration system is a heat pump with three thermal reservoirs as shown in the figure. A refrigeration effect of 100 W is required at 250 K when the heat source available is at 400 K. Heat rejection occurs at 300 K.

The minimum value of heat required (in W) is

a) 167 b) 100 c) 80 d) 20

[GATE-05]

Q.16 Dew point temperature is the temperature at which condensation begins when the air is cooled at constant a) volume b) entropy c) pressure d) enthalpy

[GATE-06] Q.17 The statements concern psychometric chart. 1. Constant relative humidity lines are uphill straight lines to the right 2. Constant wet bulb temperature lines are downhill straight lines to the right 3. Constant specific volume lines are downhill straight lines to the right 4. Constant enthalpy lines are coincident with constant wet bulb temperature Lines Which of the statements are correct ? a) 2 and 3 b) 1 and 2 c) 1 and 3 d) 2 and 4

[GATE-06]

Q.18 A building has to be maintained at 021 C (dry bulb) and 014.5 C (wet bulb).

The dew point temperature under these conditions is 010.17 C . The outside


temperature is 023 C− (dry bulb) and the internal and external surface heat transfer coefficients are 8 W/m2 K and 23 W/m2 K respectively. If the building wall has a thermal conductivity of 1.2 W/m K, the minimum thickness (in m) of the wall required to prevent condensation is a) 0.471 b) 0.407 c) 0.321 d) 0.125

[GATE-07] Q .19 Atmospheric air at a flow rate of 3 kg/s (on dry basis) enters a cooling and dehumidifying coil with an enthalpy of 85 kJ/ kg of dry air and a humidity ratio of 19 grams/kg of dry air. The air leaves the coil with an enthalpy of 43 kJ/kg of dry air and a humidity ratio of 8 grams/kg of dry air. If the condensate water leaves the coil with an enthalpy of 67 kJ/kg, the required cooling capacity of the coil in kW is a) 75.0 b) 123.8 c) 128.2 d) 159.0

[GATE-07]

Q.20 Moist air at a pressure of 100 kPa is compressed to 500 kPa and then cooled to 035 C in an aftercooler. The air at the entry to the aftercooler is unsaturated and becomes just saturated at the exit of the aftercooler. The saturation pressure of water at 035 C is 5.628 kPa. The partial pressure of water vapour (in kPa) in the moist air entering the compressor is closest to a) 0.57 b) 1.13 c) 2.26 d) 4.52

[GATE-08] Q.21 Air (at atmospheric pressure) at a dry bulb temperature of 040 C and wet

bulb temperature of 020 C is humidified in an air washer operating with continuous water recirculation. The wet bulb depression (i.e. the difference between the dry and wet bulb temperature) at the exit is 25% of that at the inlet. The dry bulb temperature at the exit of the air washer is closest to a) 010 C b) 020 C c) 025 C d) 030 C

[GATE-08]

Q.22 In an ideal vapour compression refrigeration cycle, the specific enthalpy of refrigerant (in kJ/kg) at the following states is given as: Inlet of condenser :283 Exit of condenser :116 Exit of evaporator :232 The COP of this cycle is a) 2.27 b) 2.75 c) 3.27 d)3.75

[GATE-09]

Q.23 A moist air sample has dry bulb temperature of 030 C and specific humidity of 11.5 g water vapour per kg dry air. Assume molecular weight of air as 28.93. If the saturation vapour pressure of water at

030 C is 4.24 kPa and the total pressure is 90 kPa, then the relative humidity (in %) of air sample is a) 50.5 b) 38.5 c) 56.5 d) 68.5

[GATE-10]

Q.24 If a mass of moist air in an airtight vessel is heated to a higher temperature, then a) specific humidity of the air increases b) specific humidity of the air decreases c) relative humidity of the air increases d) relative humidity of the air decreases

[GATE-11]


Q.25 The rate at which heat is extracted, in kJ/s from the refrigerated space is a) 28.3 b) 42.9 c) 34.4 d) 14.6

[GATE-12]

Q.26 The power required for the compressor in kW is a) 5.94 b) 1.83 c) 7.9 d) 39.5

[GATE-12]

Q.27 The pressure, dry bulb temperature and relative humidity of air in a room are 1 bar, 030 C and 70%, respectively. If the saturated pressure at 030 C is 4.25 kPa, the specify humidity of the room air in kg water vapour/kg dry air is a) 0.0083 b) 0.0101 c) 0.0191 d) 0.0232

[GATE-13]

Q.28 which one of the following is a CFC refrigerant ?

(a) R744 (b) R290

(C) R502 (d) R718

[GATE-2014 (1)]

Q.29 A sample of moist air at a total pressure of 85 kPa has a dry bulb temperature of 030 C (saturation vapour pressure of water = 4.24 kPa). If the air sample has a relative humidity of 65%, the absolute humidity (in gram) to water vapour per kg of dry air is_______.

[GATE-2014 (3)]

Q.30 A heat pump with refrigerant R22 is used for space heating between temperature limits of −20°C and 25°C. The heat required is 200 MJ/h. Assume specific heat of vapour at the time of discharge as 0.98 kJ/kg-K. Other

relevant properties are given below. The enthalpy (in kJ/kg) of the refrigerant at isentropic compressor discharge is _______

[GATE-2014 (2)]

Q. 31 A reversed Carnot cycle refrigerator maintains a temperature of −5°C. The ambient air temperature is 35°C. The heat gained by the refrigerator at a continuous rate is 2.5 kJ/s. The power (in watt) required to pump this heat out continuously is ____ .

[GATE-2014 (4)]

Q.32 Air in a room is at 35°C and 60% relative humidity (RH). The pressure in the room is 0.1 MPa. The saturation pressure of water at 35°C is 5.63 kPa.

The humidity ratio of the air (in gram/kg of dry air)is________.

[GATE-2015(3)]

Q.33 Refrigerant vapor enters into the Compressor of a standard vapor compression cycle at - 10oC (h = 402 kJ/kg) and leaves the compression at 50°C (h = 432 kJ/kg). It leaves the condenser at 30°C (h = 237 kJ/kg). The COP of the Cycle is ______________

[GATE-2015(3)]


Q.34 The thermodynamic cycle shown figure (T- diagram) indicates

(a) reversed Carnot cycle

(b) reversed Brayton cycle

(c) vapor compression cycle

(d) vapor absorption cycle

[GATE-2015(3)]

Q.35 The COP of a Carnot heat pump operating between 6°C and 37°C is ______

[GATE-2015(2)]

Q.36 A stream of moist air (mass flow rate 10.1 kg/s) with humidity ratio of 0.01 kg/kg dry air mixes with a second stream of superheated water vapour flowing at 0.1 kg/s. Assuming proper and Uniform mixing with no condensation, the humidity ratio of the final stream in (kg/kg dry air) is _______.

[GATE-2015(1)]

Q.37 In a mixture of dry air and water vapor at a total pressure of 750 mm of Hg, the partial pressure of water vapor is 20 mm of Hg. The humidity ratio of the air in grams of water vapor per

kg of dry air (gw/kgda) is __________. [GATE-2016 (3)]

Q.38 A refrigerator uses R-134 a as its refrigerant and operates on a ideal vapour-compression refrigeration cycle between 0.14 MPa and 0.8 MPa. If the mass flow rate of the refrigerant is 0.05 kg/s the rate of heat rejection to the environment is ____kW.

Given data :

At P = 0.14 MPa, h 236.04 kJ/kg,

s = 0.9322 kJ/kgK

At P = 0.8 MPa, h = 272.05 kJ/kg (superheated vapour)

At P = 0.8 MPa, h = 93.42 kJ/kg(saturated liquid)

[GATE-2016 (2)]

Q.39 In the vapour compression cycle shown in the figure, the evaporating and condensing temperatures are 260 K and 310K, respectively. The compressor takes in liquid-vapour mixture (state 1) and isentropically compresses it to a dry saturated vapour condition (state 2). The specific heat of the liquid refrigerant is 4.8 kJ/kgK and may be treated as constant. The enthalpy of evaporation for the refrigerant at 310 K is 1054 kJ/kg.


The difference between the enthalpies at state point 1 and 0 (in kJ/kg) is____________.

[GATE-2016 (3)]

Q.40 The partial pressure of water vapour in a moist air sample of relative humidity

70% is 1.6 kPa, the total pressure being 101.325 kPa. Moist air may be treated as an ideal gas mixture of water vapour and dry air. The relationbetween saturation temperature (Ts in K) and saturation pressure (Ps in kPa) for water is given

by ln 0

14.317 5304s

s

pp T

−=

where p0 = 101.325 kPa. The dry bulb

temperature of the moist air sample (in oC) is_________

[GATE-2016 (2)]

Q.41 Moist air is treated as an ideal

gas mixture of water vapor and dry air

(molecular weight of air = 28.84 and

molecular weight of water = 18). At a

location, the total pressure is 100 kPa,

the temperature is 30°C and the relative

humidity is 55%. Given that the

saturation pressure of water at 30°C is

4246 Pa, the mass of water vapor per kg

of dry air is ______ grams.

[GATE-2017 (1)]

Q.42 If a mass of moist air contained in a closed metallic vessel is heated, then its

a) relative humidity decreases

b) relative humidity increases

c) specific humidity increases

d) specific humidity decreases

[GATE-2017 (2)]

Q.43 Ambient air is at a pressure of 100 kPa, dry bulb temperature of 30 0 C and 60% relative humidity. The saturation pressure of water at 30 0 C is 4.24 kPa. The specific humidity of air (in g/kg of dry air) is ________ (correct to two decimal places).

[GATE-2018 (1)]

Q.44 A standard vapor compression refrigeration cycle operating with a condensing temperature of 35 0 C and an evaporating temperature of -10 0 C develops 15 kW of cooling.

The p-h diagram shows the enthalpies at various states. If the isentropic efficiency of the compressor is 0.75, the magnitude of compressor power (in kW) is _________ (correct to two decimal places).

[GATE-2018 (2)]


1 2 3 4 5 6 7 8 9 10 11 12 13 (c) (b) (a) (b) (a) (b) (d) (c) (b) (a) (b) (b) (a) 14 15 16 17 18 19 20 21 22 23 24 25 26 (b) (c) (c) (a) (b) (c) (b) (c) (a) (b) (d) (a) (c) 27 28 29 30 31 32 33 34 35 36 37 38 39 (c)

(b)

20.84

(b)

433.3

373.13

21.74

5.5

(b)

10

0.02

17.04

8.93

27

28

29

30

31

32

1103.51

19.89

14.9

(a)

16.236

10

ANSWER KEY:


Q.1(c)

1H.P.

1 2

Q 1000So, (COP) = 4Q Q 1000 750

= =− −

Q.2 (b) We know that for saturated air, the relative humidity is 100% and the dry bulb temperature, wet bulb temperature and dew point temperature is same. But when air is not saturated, dew point temperature is always less than the wet bulb temperature. DPT < WBT

Q.3(a)

EXPLANATIONS


01 4

02 3

02 2

3 4

Given : T = T =- 20 C = (- 20 + 273)K = 253 K,

m = 0.025 kg/ secT =T = 40 C = (40 + 273)K = 313 KFrom the given table,At, T = 40 C, h = 200 kJ/kgAnd h = h = 80 kJ/kgFrom the given T s - e

curv

•

1 2

2 f fg

2 20

s = ss = s + xs

{s is taken 0.67 because s at the temperature 40 C & at 2 high temperatureand pressure vapour refrigerant exist.}0.67 = 0.07 + x(0.7366 - 0.07)x=0.90

1 f fg

1

And Enthalpy at point 1 is,h = h + xh = 20 + 0.90(180 - 20) = 164 kJ/kgNow refrigeration effect is produce in the evaporator.Heat extracted from the evaporator or refrigerating effect,

RE = m (h•

4- h ) = 0.025(164 - 80) = 2.1 kW

Q.4 (b)

1 4

2 1

Refrigerating effectCOP)refrigerator Work done

h h 164 80= = = 2.33h h 20

(

0 16

=

− −− −

[GATE-03]

Q.5 (a)Air conditioner mounted in a window or through the wall are self-contained units of small capacity of 1 TR to 3 TR. The capillary tube is used as an expansion device in small capacity refrigeration units Q.6 (b)

In the process of chemical dehumidification of air , the air is passed over chemicals which have an affinity for moisture and the moisture of air gets condensed out and gives up its latent heat. Due to the condensation, the specific humidity decreases and the heat of condensation supplies sensible heat for heating the air and thus increasing its dry bulb temperature. So chemical dehumidification increase dry bulb temperature & decreases specific humidity Q.7 (d) If a refrigerant is written in the form of Rabc. The first digit on the right (c) is the number of fluorine (F) atoms, the second digit from the right (b) is one more than the number of hydrogen (H) atoms required & third digit from the right (a) is one less than the Number of carbon (C) atoms in the refrigerant. So, For R134 First digit from the Right = 4 = Number of Fluorine atoms Second digit from the right = 3 - 1 = 2 = Number of hydrogen atoms Third digit from the right = 1 + 1 = 2 = Number of carbon atoms Hence, Chemical formula is C2H2F4


Q.8 (c)

( ) ( )

( )

( )

( ) ( )

H.E

3

H.E1

3

1

3

1

COP 5, η 70% 0.7

COP 5........( )

η 0.7..........( )

Multiplying i and ii

5 0.7

3.5

ref

ref

Q iW

W iiQ

Q WW QQQ

= = =

= =

= =

= ×

=

Hence, Energy absorbed (Q3) from low temperature reservoir by the refrigerator for each kJ of energy absorbed (Q1) from high temperature source by the engine= 3.5 kJ

Q.9(b) Given : tdp = 180C = (273 + 18)K = 291K,

p = patm = 1.013 bar tdb = 300C = (273 + 30)K = 303 K pv = 0.02062 bar (for water vapour at dew point). cair = 1.005 kJ/kg K, cwater = 1.88 kJ/kg K Latent heat of vaporization of water at 00C. hfgdp = 2500 kJ/kg

Specific humidity, 0.622

kg 0.01291kg of dry air

v

v

pWp p

W

= ×−

=

( )db fgdp

Enthalpy,

h 1.022 T h 2.3

h 63.15

= + +

=

dpt

KJkg

ω

Q.10 (a) Q.11 (b) From steam table, saturated air pressure corresponding to dry bulb temperature of 350C is ps = 0.05628 bar. Relative humidity

v

vs

vv

vs

v

b v

PP

P0.75 P 0.04221 barP

pSpecific humidity = 0.622p p

0.04221= 0.6221.01 0.04221

Rela

0.0271 kg/kg of dry ai

tive humi t

r

di y φ =

= ⇒ =

ω−

ω − =

Q.12 (b)Given :

Hence air gets heated, Also water is added to it, so it gets humidified.

0 0 0sp db wb

sp db

t 42 C, t 40 C, t 20 CHere we see that t t

= = =>

Q.13 (a)Given curve is the theoretical p-h curve for vapour compression refrigeration cycle. Q.14(b)

Q.15(c)

Q.16(c)


It is the temperature of air recorded by a thermometer, when the moisture (water vapour) present in it begins to condense. If a sample of unsaturated air, containing superheated water vapour, is cooled at constant pressure, the partial pressure (pv) of each constituent remains constant until the water vapour reaches the saturated state as shown by point B. At this point B the first drop of dew will be formed and hence the temperature at point B is called dew point temperature Q.17(a)

Q.18 (b)

Let h1 & h2 be the internal and external surface heat transfer coefficients respectively and building wall has thermal conductivity k . Given : h1= 8 W/m K h2 =23W/m2 K, k=1.2 W/m K, TDPT=10.17 0C Now to prevent condensation, temperature of inner wall should be more than or equal to the dew point temperature. It is the

limiting condition to prevent condensation Ts1=10.17 0C Here Ts1 & Ts2 are internal & external wall surface temperature of building. Hence, heat flux per unit area inside the building,

( )1 1 1

286.64 W/m

i DBT s

i

Qq h T TA

q

= = −

=…….(i)

Heat flux per unit area outside the building is

q0 = h2(Ts2- TDBT2) = 23(Ts2+ 23) …....(ii) Heat flow will be same at inside & outside the building. So from equation (i) & (ii) qi = q0 Ts2 = -19.23 0C For minimum thickness of the wall, use the Fourier’s law of conduction for the building. Heat flux through wall,

( )1 2

0.407 m

s sk T Tq

xx

−=

=

Q.19 (c) Given : ma = 3 kg/sec,

Using subscript 1 and 2 for the inlet and outlet of the coil respectively. h1 = 85 kJ/kg of dry air, W1 = 19 grams/kg of dry air =19 ×10-3 kg/kg of dry air h2 = 43 kJ/kg of dry air, W2 = 8 grams/kg of dry air = 8 ×10-3 kg/kg of dry air h3 = 67 kJ/kg Mass flow rate of water vapour at the inlet of the coil is, mv1 = W1 ×m1 = 57 ×10-8 kg/sec And mass flow rate of water vapour at the outlet of coil is, mv2 = W2 ×ma


mv2 =24 ×10-8 kg/sec So, mass of water vapour condensed in the coil is, Therefore, required cooling capacity of the coil = change in enthalpy of dry air + change in enthalpy of condensed water = (85 - 43) ×3 + 67 ×33 ×10-3 = 128.211 kW

Q.20 (b) Given : p1 = 100 kPa, p2 = 500 kPa, pv1 = ? pv2 = 5.628 kPa (Saturated pressure at 350C) We know that,

8


For CASE-II0.6280.622

500 5.6287.08 10 /

v

v

pWp p

W

W kg kg of dry air−

= ×−

= − = ×

For saturated air specific humidity remains same. So, for case (I) :

1

1 1

0.622 v

v

pWp p

= ×−

On substituting the values, we get

1 1.13 vp kPa=

Q.21 (C)

Given: At inlet tDBT =40 0C tWBT=20 0C We know that, Wet bulb depression = tDBT -tWBT

= 40 - 20 = 20 0C And given wet bulb depression at the exit = 25% of wet bulb depression at inlet This process becomes adiabatic saturation and for this process,

tWBT(inlet) = tWBT(outlet) So, tDBT(exit)- 20 = 0.25*20 tDBT(exit) = 20 + 5 = 25 0C

Q.22 (a) p-h curve for vapour compression refrigeration cycle is as follows

The given specific enthalpies are Inlet of condenser h2 = 283 kJ/kg

Exit of condenser h3 = 116 kJkg

=h4 (from p-h curve)

Exit of evaporator h1 = 232 kJ/kg Now,

1 4

2 1

Refrigeration EffectCOP=Work Done

2.27h hh h−

= =−

Q.23 (b)

Given : tDBT = 30 0C, W = 11.5 g water vapour/kg dry

air ps = 4.24 kPa, p = 90 kPa


8


11.5 10 0.62290

1.634 kPa

v

v

v

v

v

pWp p

pp

p

−

= ×−

× = − =

( )Relative humidity,

1.634 4.24

0.385 38.5%

v

s

pp

φ

φ

=

=

= =

Q.4 (d) From the given curve, we easily see that relative humidity of air decreases, when temperature of moist air in an airtight vessel increases. So, option (d) is correct. Specific humidity remain constant with temperature increase, so option a & b are incorrect

A refrigerator operates between 120 kPa and 800 kPa in an ideal vapour compression cycle with R-134a as the refrigerant. The refrigerant enters the compressor as saturated vapour and leaves the condenser as saturated liquid. The

mass flow rate of the refrigerant is 0.2 kg/s. Properties for R134a are as follows :

Q.25 (a)

T-S

T-S Diagram for given Refrigeration cycle is given above Since Heat is Extracted in Evaporation process. So rate of heat extracted = ( )1 4m h h− From above diagram (h3 = h4) for throttling process, so Heat extracted = m (h1- h3) From given table h1 = hg at 120 kPa, hg = 237 kJ/kg h3 = hf at 120 kPa, hf = 95.5 kJ/kg Hence Heat extracted = m (hg - hf) = 0.2(237 - 95.5) = 28.3 kJ/s Q.26 (c) Since power is required for compressor in refrigeration is in compression cycle (1-2) Hence, Power required = h2-h1 = h2-hf Since for isentropic compression process.


s1 = s2 from figure. = 0.95 For entropy s = 0.95 the enthalpy h = 276.45 kJ/kg h = h2 = 276.45 (From table) Hence Power = 0.2(276.45 - 237) = 7.89 = 7.9 kW Q.27 (c)

Specific Humidity is given by

0.622 v

a v

pp p

ω = ×−

…...(i)

Where, Relative Saturated

humidity steam pressure

0.7 0.0425 0.02975 bar

v

s

p

pφ

= ×

= × = ×=

So that from equation (i), we have

0.029750.622 1 bar1 0.02975

0.0191 kg/kg of dry air

aPω = × = −=

Q.28

( )2

3 3

3 2 3

R744-COR290-C H Propane

R502-CHCIF CCIF CFR718-Water

+

Q.29

odb

s

Total pressure : p = 85kPaDry bulb temperature,T 30 Cp 4.24kPa

65 % 0.65Relative humidity,

0.654.24

0.65 4.24 2.756 kPa

v

s

v

v

pp

p

or p

φ

φ

=

== =

=

=

= =

Absolute humidity,0.622=

0.02084 kg of w.v/kg of dry air= 20.84 gram of w.v./kg of dry air

v

v

pp p

ω−

=

2 1 Dehumidifierω < ω Q

Q.30

( )

22 2 p e

2

1 2

2

2

p

2 2 2e

2 p

2e

2

Tc logT

for isentropic process

1.7841 kJ/kgKs 1.7183 kJ/kgKc 0.98 kJ/kgK

TlogT c

T 1.7841 1.7183log 0.0671T 0.98

− =

=

=

=

=

−∴ =

−= =

''

'

'

'

s s

s s

s

s s


( )( )

2 2

2 2 p 2 2

2

2

or T 1.0694 T 1.0694 298318.695K

h h c T T

h 413.02 0.98 318.695 298

h 433.3 kJ/kg

= ==

− = −

= + −

=

'

' '

Q.31

Given data :

( )( )

2

1

T 5 C 5 273 K = 268 K

T 35 C 35 273 K 308 K

= − ° = − +

= ° = + =

2Q 2.5kJ/s = 2.5 kW = 2500 W=

( )

( )

( )

2R

1 2

R

2

TCOPT T

268COP 6.7308 268

Qalso COPW

25006.7W2500or W = 373.13 watt6.7

=−

= =−

=

∴ =

=

Q.32

Given data :

Tdb = 35°C

φ = 60 % = O.60

p = 0.1 MPa = 100 kPa

ps = 5.63 kPa at 35°C

Relative humidity,

0.605.630.60 5.63 3.378 kPa

Humidiy ratio,0.622 0.622 3.378=

100 3.378

v

s

v

v

v

v

pp

p

p

pp p

φ =

= =

ω =− −

= 0.02174 kg of w.v./kg of dry air

= 21.74 gram of w. v./kg of dry air

Q.33

1 2

3 4

Given data :402kJ/kg, h 432kJ/kg

h h 237 kJ/kgh = =

= =

1 4

2 1

402 237 165COP =432 402 30

5.5

h hh h− −

= =− −

=

Q.34 (b)

Reversed Brayton cycle is shown in the figure


Q.35

( )( )

o2

o1

Given data :T 6 C = 6 273 K= 279K

T 37 C = 37 273 K= 310K

= +

= +

( ) 1HP

1 2

T 310COPT T 310 279

10

= =− −

=

Q.36

Mass flow rate of moist air = 10.1 kg/s

Humidity ratio : v

a

m 0.01 kg/sm

ω = =

Mass of moist air = Mass of dry air + Mass of water vapour

a a

a

10.1 m 0.01 m10.1 m 10.1

= +

=

Mass of dry air a10.1m = = 10 kg/s1.01

Mass of water vapour,

( )

v1

v2

v v1 v2total

vfinal

a

m 10.1 10 0.1 kg/sm 0.1 kg/s

m m m 0.2 kg/s

m 0.2Humidity ratiom 10

= − =

=

= + =

ω = =

= 0.02 kg/kg of dry air

Q.37

Given data :

Total pressure,

p = 750mm of Hg

Partial pressure, V

Pv = 20mm of Hg

We know that humidity ratio,

0.622 kg. w.v./kg d.a.

0.622 20=750 20

0.01704 kg w.v./kg of d.a.17.04 g of w.v./kg of d.a.

v

v

pp p

ω =−

−==

Q.38

Given data:

m = 0.05kg/s

Pe = 0.14 MPa

h1 = 236.04 kJ/kg


S1 = 0.9322 kJ/kgK

p = 0.8 MPa

h3 = h4 = hf = 93.42 kJ/kg

h2 = 272.02 kJ/kg

Rate of heat rejection to the environment,

[ ]2-3 2 3Q 0.05 275.02 93.42

8.93 kW

m h h = − = − =

Q.39

Given data :

Te = 260K

TC = 310K

cpl = 4.8 kJ/kgK

hfg = 1054 kJ/kg at TC = 310 K

Taking reference temperature T, at which entropy

is ST.

33 T

33 T

T

0 T

T

fg2 3

2 3

3

2 1

1 T

Ts s c lnT

Ts s c lnT

310 s 4.8 lnT

Similarly, 260s s c lnT

260 s 4.8 lnT

h 1054s sT 310

1054s s310

Substitute the value of s in above equation, we gets s

310 1054s s 4.8 lnT 310

h

∴ − =

= +

= +

= +

= +

∴ − = =

= +

=

= + +

∴ 1 0 1 0

310 1054 260260 4.8 ln 4.8lnT 310 T

310 T 1054260 4.8lnT 260 310

1103.51 kJ/kg

e

T T

h T s s

s s

− = − = + + − −

= + =

Q.40

Given data :

Relative humidity,

φ = 70 % = 0.70

Partial pressure,

pv = 1.6 kPa

Total pressure,

p0 = 101.325 kPa

We know that,


v

s

pp

φ =

1.60.70

1.6 2.2857 kPa0.70

s

s

p

or p

=

= =

Temperature corresponding to saturation pressure

is dry bulb temperature,

( )

s

s

s

s

s

so

o

5304ln 14.317T

2.2857 5304ln 14.317101.325 T

53043.79166 14.317T

5304 14.317 3.79166T

5304 18.10866T

T 292.89K= 292.89 273 C

= 19.89 C

s

o

pp

or

= −

= −

− = −

= +

=

=

−

Q.41

Relative humidity v

vs

PP

φ =

vP0.554.246

=

vP 2.335=

2H O v

air t v

M PM P P

ω = −

18 2.335

28.84 100 2.335 ω = −

Kg of water vapour0.0149

Kg of dry airω =

v

a

mm

ω =Q

v am m= ω

30.0149 10 1= × ×

vm 14.9grm=

Q.42 (a)

For closed metallic vessel sp. Humidity (ω) remain const

From the figure

2 1φ < φ

Q.43

v

s

a s

v

b v

b

PRelative humidity =P

But, 60 % 0.6 , 4.24 kPap p 0.6 4.24 2.544 kPa

pSpecific humidity = 0.622p p

where p ambient air pressure = 100 kPa2.5440.622

100 2.5440.016236kg/kg of dry air

= 1

= = =

⇒ = = =

ω−

=

⇒ ω = − ⇒ ω =⇒ω

sP

φ

φ

φ

6.236 g/kg of dryair


Q.44 (10)

( )( )( )

( ) ( )

1 4

isentropic 2 1

isentropicC

actual

actual actual

m

RC = 15 kW

RC = m

RC = m 400 250

kg/sec

475 400 75kJ/kg

η

m 0.1 10010 kWP

h h

m

W h h

WW

W W

−

−

= − = − =

=

= =

=.

.

.

.

.


thermodynamics · 2.7 first law of thermodynamics for variable flow process 17 2.8 examples 18 3....

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